Math Ncert Exemplar 2019 for Class 9 Maths Chapter 8 - Quadrilaterals

Answer:

Given that, $\angle \mathrm{A}=75°, \angle \mathrm{B}=90° \mathrm{and} \angle \mathrm{C}=75°$.

The sum of all the angles of a quadrilateral is 360°.
$$\begin{gathered} \therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360° \\ \Rightarrow 75°+90°+75°+\angle \mathrm{D}=360° \end{gathered}$$
$\begin{array}{rcl}& \Rightarrow & \angle \mathrm{D}=360°-\left(75°+90°+75°\right)\\ & =& 360°-240°\\ & =& 120°\end{array}$
Thus, the fourth angle of the quadrilaterals is 120°.
Hence, the correct answer is option D.

Answer:

The diagonals of rectangle are equal in length.

$\therefore \mathrm{AC}=\mathrm{BD}$
$$\begin{gathered} \Rightarrow \frac{1}{2} \mathrm{AC}=\frac{1}{2} \mathrm{BD} \\ \Rightarrow \mathrm{OA}=\mathrm{OB} \left(\because \mathrm{O} \mathrm{is} \mathrm{the} \mathrm{midpoint} \mathrm{of} \mathrm{AC} \mathrm{and} \mathrm{BD}\right) \end{gathered}$$
$\begin{array}{rcl}& \Rightarrow & \angle 2=\angle 1 \left(\because \mathrm{Angles} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}\right)\\ & =& 25°\end{array}$
$\begin{array}{rcl}\therefore \angle 3& =& \angle 1+\angle 2 \left(\because \mathrm{Exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{sum} \mathrm{of} 2 \mathrm{opposite} \mathrm{interior} \mathrm{angles}\right)\\ & =& 25°+25°\\ & =& 50°\end{array}$
Thus, the acute angle between diagonals is 50°.
Hence, the correct answer is option B.

Answer:

Given, ABCD is a rhombus,
$$\begin{gathered} \angle \mathrm{ACB}=40° \\ \Rightarrow \angle \mathrm{OCB}=40° \end{gathered}$$

As, $\mathrm{AD}\parallel \mathrm{BC}$
$\angle \mathrm{DAC}=\angle \mathrm{BCA}=40° \left[\because \mathrm{alternate} \mathrm{interior} \mathrm{angle}\right]$

Also, 
$\angle \mathrm{AOD}=90° \left[\because \mathrm{diagonals} \mathrm{of} \mathrm{rhombus} \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other}\right]$

By angle sum property of triangles, the sum of all angles of $△$ADO is 180­°.
$\therefore \angle \mathrm{ADO}+\angle \mathrm{DOA}+\angle \mathrm{OAD}=180°$
$\begin{array}{rcl}& \Rightarrow & \angle \mathrm{ADO}=180°-\left(40°+90°\right)\\ & =& 180°-130°\\ & =& 50°\end{array}$
$\Rightarrow \angle \mathrm{ADB}=\angle \mathrm{ADO}=50°$
Hence, the correct answer is option C.

Answer:

Rectangle ABCD is formed by joining the mid points of quadrilateral PQRS.


Applying midpoint theorem in $△$PQR and $△$PSR,
AB = CD = $\frac{1}{2}\mathrm{PR}$
and
AB ∥ PR ∥ CD

Similarly, applying midpoint theorem in $△$PQS and $△$QRS,
AD = BC = $\frac{1}{2}\mathrm{QS}$
and
AD ∥ QS ∥ BC

Since, QS ∥ BC therefore,
∠OXB + ∠XBY = 180°            (∵ sum of co-interior angles is 180°)        .....(1)

Now, ABCD is a rectangle.
∴ ∠XBY = 90°             .....(2)         

Putting (2) in (1),
 ∠OXB = 90°              .....(3)

Again, AB ∥ PR.
⇒ BX ∥ OY
Therefore,
∠OXB + ∠XOY = 180°            (∵ sum of co-interior angles is 180°)        .....(4)

Using (3),
∴ ∠XOY = 90°
Thus, PR is perpendicular to QS.

Therefore, the diagonals of PQRS are perpendicular to each other.
Hence, the correct answer is option C.
 

Answer:

Given, quadrilateral ABCD is a rhombus.
So, AB = BC = CD = AD.

In $△$PQS,
D and A are the mid points of PS and PQ respectively.
So, by mid point theorem,
AD = BC = $\frac{1}{2}$QS           .....(1)
and
AD ∥ QS ∥ BC

Similarly in $△$PSR,
D and C are the mid points of PS nad RS respectively.
So, by mid point theorem,
DC = AB = $\frac{1}{2}$PR                 .....(2)
and
DC ∥ PR ∥ AB

Since, ABCD is a rhombus.
Therefore, all its sides are equal.
⇒ AB = BC

Using (1) and (2),
$\frac{1}{2}$QS = $\frac{1}{2}$PR
$\Rightarrow \mathrm{QS}=\mathrm{PR}$
Thus, diagonals of PQRS are equal.
Hence, the correct answer is option D.
 

Answer:

Given that, the angles of quadrilaterals in ratio 3 : 7 : 6 : 4.

Let the angles be $3x, 7x, 6x \mathrm{and} 4x.$
∴ Sum of all angles of quadrilaterals = 360°
$$\begin{gathered} \Rightarrow 3x+7x+6x+4x=360° \\ \Rightarrow 20x=360° \\ \Rightarrow x=18° \end{gathered}$$

Now, the angles of quadrilaterals are
$$\begin{gathered} \angle \mathrm{A}=3\times 18=54° \\ \angle \mathrm{B}=7\times 18=126° \\ \angle \mathrm{C}=6\times 18=108° \\ \angle \mathrm{D}=4\times 18=72° \end{gathered}$$

From the figure,
$$\begin{gathered} \angle \mathrm{BCE} = 180°-\angle \mathrm{BCD} \left[\mathrm{Linear} \mathrm{pair} \mathrm{axiom}\right] \\ \Rightarrow \angle \mathrm{BCE}=180°-108°=72° \end{gathered}$$

Here, $\angle \mathrm{BCE}=\angle \mathrm{ADC}=70°$.

Since, corresponding angle are equal.
$\therefore \mathrm{BC}\parallel \mathrm{AD}$

Then, sum of co-interior angles,
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}=126°+54°=180° \\ \angle \mathrm{C}+\angle \mathrm{D}=180°+72°=180° \end{gathered}$$
Thus, ABCD is a trapezium.

Hence, the correct answer is option C.
 

Answer:

Given that, ABCD is a quadrilateral and all angles bisectors from a quadrilateral PQRS.

The sum of all angles in a quadrilateral is 360°.
$$\begin{gathered} \therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360° \\ \Rightarrow \frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}\right)=180° \\ \Rightarrow \angle \mathrm{PAB}+\angle \mathrm{PBA}+\angle \mathrm{RCD}+\angle \mathrm{RDC}=180° .....\left(1\right) \end{gathered}$$

In $△$APB, by angle sum property,
$$\begin{gathered} \angle \mathrm{PAB}+\angle \mathrm{ABP}+\angle \mathrm{BPA}=180° \\ \Rightarrow \angle \mathrm{PAB}-\angle \mathrm{ABP}=180°-\angle \mathrm{BPA} .....\left(2\right) \end{gathered}$$

Similarly, in $△$RDC,
$$\begin{gathered} \angle \mathrm{RDC}+\angle \mathrm{DCR}+\angle \mathrm{CRD}=180° \\ \Rightarrow \angle \mathrm{RDC}+\angle \mathrm{DCR}=180-\angle \mathrm{CRD} .....\left(3\right) \end{gathered}$$

Substituting (2) and (3) in (1),
$$\begin{gathered} 180°-\angle \mathrm{BPA}+180°-\angle \mathrm{DRC}=180° \\ \Rightarrow \angle \mathrm{BPA}+\angle \mathrm{DRC}=180° \\ \Rightarrow \angle \mathrm{SPQ}+\angle \mathrm{SRQ}=180° \end{gathered}$$
[∵$\angle \mathrm{BPA}=\angle \mathrm{SPQ}$ and $\angle \mathrm{DRC}=\angle \mathrm{SRQ}$ are vertically opposite angles]
Thus, PQRS is a quadrilateral whose opposite angles are supplementary.
Hence, the correct answer is option D.
 

Answer:

Given that, APB and CQD are two parallel lines.
The bisector of the angles APQ and CQP meet at point M.
Similarly, the bisectors of angles BPQ and PQD meet at point N.
Join PM, MQ, QN and NP.


Since, $\mathrm{AB}\parallel \mathrm{CD}$.
Then, $\angle \mathrm{APQ}=\angle \mathrm{PQD} \left[\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right]$
$$\begin{gathered} \Rightarrow 2\angle \mathrm{MPQ}=2\angle \mathrm{NQP} \\ \Rightarrow \angle \mathrm{MPQ}=\angle \mathrm{NQP} \end{gathered}$$
$\therefore \mathrm{PM}\parallel \mathrm{QN}$

Similarly, $\angle \mathrm{BPQ}=\angle \mathrm{CQP} \left[\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right]$
$\therefore \mathrm{PN}\parallel \mathrm{QM}$

So, quadrilateral PMQN is a parallelogram.
$$\begin{gathered} \therefore \angle \mathrm{CQD}=180° \left[\mathrm{Since}, \mathrm{CQD} \mathrm{is} \mathrm{a} \mathrm{line}\right] \\ \Rightarrow \angle \mathrm{CQP}+\angle \mathrm{DQP}=180° \\ \Rightarrow 2\angle \mathrm{MQP}+2\angle \mathrm{NQP}=180° \\ \Rightarrow \angle \mathrm{MQP}+\angle \mathrm{NQP}=90° \\ \Rightarrow \angle \mathrm{MQN}=90° \end{gathered}$$
Thus, PMQN is a rectangle.
Hence, the correct answer is option C.

Answer:

Let ABCD be a rhombus in which P, Q, R, and S are the mid-points of side AB, BC, CD and DA respectively.

Join AC, PR and SQ.
In $△$ABC,
P is the mid-point of AB and Q the mid-point of BC.
$\Rightarrow \mathrm{PQ}\parallel \mathrm{AC} \mathrm{and} \mathrm{PQ}=\frac{1}{2}\mathrm{AC} \left[\mathrm{using} \mathrm{midpoint} \mathrm{theorem}\right] .....\left(1\right)$

Similarly, in $△$DAC,
$\mathrm{SR}\parallel \mathrm{AC} \mathrm{and} \mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(2\right)$$$

From (1) and (2),
$\mathrm{PQ}\parallel \mathrm{SR} \mathrm{and} \mathrm{PQ}=\mathrm{SR}$

So, PQRS is a parallelogram.
$$\begin{gathered} \Rightarrow \mathrm{BC}=\mathrm{PR} \left[\mathrm{opposite} \mathrm{side} \mathrm{of} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}\right] \\ \Rightarrow \mathrm{AB}=\mathrm{PR} \left[\because \mathrm{BC}=\mathrm{AB} \mathrm{are} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{rhombus}\right] \\ \Rightarrow \mathrm{SQ}=\mathrm{PR} \left[\mathrm{From} \left(1\right)\right] \end{gathered}$$
So, the diagonals of a parallelogram are equal.

Thus, PQRS is a rectangle.
Hence, the correct answer is option B.

Answer:

In $△$ABC, D and E are the mid-point of sides AB and AC, respectively.

By mid-point theorem,
$\mathrm{DE}\parallel \mathrm{BC} \mathrm{and} \mathrm{DE}=\frac{1}{2}\mathrm{BC} .....\left(1\right)$
Now, P and Q are the mid-points of OB and OC.
$$\begin{gathered} \mathrm{DE}=\frac{1}{2}\left[\mathrm{BP}+\mathrm{PO}+\mathrm{OQ}+\mathrm{QC}\right] \\ \Rightarrow \mathrm{DE}=\frac{1}{2}\left[2\mathrm{PO}+2\mathrm{OQ}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{DE}=\mathrm{PO}+\mathrm{OQ} \\ \Rightarrow \mathrm{DE}=\mathrm{PQ} .....\left(2\right) \end{gathered}$$

Again, in $△$AOC, Q and E are the mid-points of OC and AC.
$\therefore \mathrm{EQ}\parallel \mathrm{AO} \mathrm{and} \mathrm{EQ}=\frac{1}{2}\mathrm{AO}$  [mid-point theorem]            .....(3)
And, in $△$ABO,
$\mathrm{PD}\parallel \mathrm{AO} \mathrm{and} \mathrm{PD}=\frac{1}{2}\mathrm{AO} .....\left(4\right)$  [mid-point theorem]

From (1) and (2),
DE$\parallel$BC (or DE$\parallel$PQ) and DE = PQ

From (3) and (4),
EQ$\parallel$PD and EQ = PD

Thus, DEQP is a parallelogram.
Hence, the correct answer is option D.

Answer:

Given that, ABCD is a quadrilateral and P, Q, R and S are the mid-points of sides of AB, BC, CD and DA.

 Also, PQRS is a square.
∴ PQ = QR = RS = PS        .....(1)
Now, the diagonals of a square are equal. Thus, PR = SQ.
But PR = BC and SQ = AB
∴ AB = BC
Therefore, all the sides of quadrilateral ABCD are equal.

Now, quadrilateral ABCD is either a square or a rhombus.

In $△$ADB, by mid-point theorem
$\mathrm{SP}\parallel \mathrm{DB} \mathrm{and} \mathrm{SP}=\frac{1}{2}\mathrm{DB} .....\left(2\right)$
Now, in $△$ABC, (by mid-point theorem)
$\mathrm{PQ}\parallel \mathrm{AC} \mathrm{and} \mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(3\right)$

From (1),
$$\begin{gathered} \mathrm{PS}=\mathrm{PQ} \\ \Rightarrow \frac{1}{2}\mathrm{DB}=\frac{1}{2}\mathrm{AC} \left[\mathrm{from} \left(1\right) \mathrm{and} \left(2\right)\right] \\ \Rightarrow \mathrm{DB}=\mathrm{AC} \end{gathered}$$ 
Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square not rhombus, So, diagonals of quadrilateral are also perpendicular.
Hence, the correct answer is option C.

Answer:

Given that, $\angle \mathrm{AOB}=70° \mathrm{and} \angle \mathrm{DAC}=32°$.

$\therefore \angle \mathrm{ACB}=32° \left[\because \mathrm{AD}\parallel \mathrm{BC} \mathrm{and} \mathrm{AC} \mathrm{is} \mathrm{transversal}\right]$

Now, $\angle \mathrm{AOB}+\angle \mathrm{BOC}=180°$           [linear pair axiom]
$\begin{array}{rcl}& \Rightarrow & \angle \mathrm{BOC}=180°-\angle \mathrm{AOB}\\ & =& 180°-70°\\ & =& 110°\end{array}$

Now, in $△$BOC,
$\angle \mathrm{BOC}+\angle \mathrm{BCO}+\angle \mathrm{OBC}=180°$      [angle sum property of a triangle]
$\Rightarrow 110°+32°+\angle \mathrm{OBC}=180°$           [$\therefore \angle \mathrm{BCO}=\angle \mathrm{ACB}=32°$]
$\begin{array}{rcl}& \Rightarrow & \angle \mathrm{OBC}=180°-\left(110°+32°\right)\\ & =& 38°\\ \therefore \angle \mathrm{DBC}& =& \angle \mathrm{OBC}=38°\end{array}$

Hence, the correct answer is option C.
 

Answer:

In a parallelogram opposite sides are equal, opposite angles are equal, diagonal bisect each other, but opposite angles are not bisected by the diagonal.
Hence, the correct answer is option C.

Answer:

In $△$ADE and $△$CFE, assume DE = EF as E is the mid-point of DF.
Now,  AE = CE.


Let DE = EF
and $\angle \mathrm{AED} = \angle \mathrm{FEC}$  [vertically opposite angles]
$\therefore △\mathrm{ADE}\cong △\mathrm{CFE}$     [SAS congruence rule]
⇒ AD = CF                   [CPCT rule]

Also, $\angle \mathrm{ADE}=\angle \mathrm{CFE}$       [CPCT]
Hence, $\mathrm{AD}\parallel \mathrm{CF}$                [alternate interior angles are equal]

Therefore, we need an additional information which is DE = EF.
Hence, the correct answer is option C.

Answer:

Given that, ABCD is a parallelogram. Also, OA = 3 cm and OD = 2 cm.

The diagonals of a parallelogram bisects each other.
∴ Diagonal AC = 2 × OA = 6 cm
and Diagonal BD = 2 × OD = 4 cm
Hence, the length of the diagonal AC and BD are 6 cm and 4 cm respectively.

Answer:

No, diagonals of a parallelogram are not perpendicular to each other, but they bisect each other.

Answer:

As, the sum of the angles of the quadrilateral is 360°.
Here,
110° + 80° + 70° + 95° = 355°$\ne$ 360°

Hence, the angles cannot be the angles of a quadrilateral.

Answer:

As, the sum of adjacent opposite angle is 180°. Thus, the given quadrilateral is a trapezium.

Answer:

The sum of all the angles of a quadrilateral is 360°.

If ABCD is a quadrilateral,
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°$

If all angles are equal, i.e.,
$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}$
$$\begin{gathered} \Rightarrow 4\angle \mathrm{A}=360° \\ \Rightarrow \angle \mathrm{A}=90° \end{gathered}$$
$\therefore \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90°$

As, all the angles are 90°.
Hence, the given quadrilateral is either a rectangle or a square.

Answer:

No, diagonals of a rectangle are equal but need not be perpendicular to each other.

Answer:

No, all the 4 angles of a quadrilateral cannot be obtuse. This is because the sum of the angles of a quadrilateral is 360°.
At max 3 obtuse angles can exist.

Answer:

In $△$ABC,
AB = 5 cm, BC = 8 cm and CA = 7 cm

As, D and E are mid-points of AB and BC.

By mid-point theorem,
$\mathrm{DE}\parallel \mathrm{AC}$
and
$\mathrm{DE}=\frac{1}{2}\mathrm{AC}=\frac{7}{2}=3.5 \mathrm{cm}$

Hence, the length of DE = 3.5 cm.

Answer:

Yes, in the given figure, BDEF is a parallelogram.
$\therefore \mathrm{BD}\parallel \mathrm{EF} \mathrm{and} \mathrm{BD}=\mathrm{EF} .....\left(1\right)$

Also, FDCE is a parallelogram.
$\therefore \mathrm{CD}\parallel \mathrm{EF} .....\left(2\right)$

From (1) and (2),
BD = CD = EF
Hence, we can say BD = CD.

Answer:

Given that, ABCD and AEFG are two parallelograms.
Also, $\angle \mathrm{C}=55°$.

As ABCD is a parallelogram, then its opposite angles are equal.
$$\begin{gathered} \Rightarrow \angle \mathrm{A}=\angle \mathrm{C}=55° \\ \Rightarrow \angle \mathrm{A}=55° \end{gathered}$$

Also, AEFG is a parallelogram.
$\therefore \angle \mathrm{A}=\angle \mathrm{F}=55°$

Hence, $\angle \mathrm{F}=55°$.

Answer:

No.

The sum of the angles of quadrilateral is 360°.

all the angles of a quadrilateral cannot be acute angles. In this case, the sum of the angles will be less than 360°. Thus, a maximum 3 acute angles are possible.

Answer:

Yes, all the angles of a quadrilateral can be right angles. In this case, the quadrilateral is rectangle or a square.

Answer:

As, the diagonals of a quadrilateral bisect each other, thus it can be a parallelogram.

Therefore, the sum of co-interior angles between two parallel lines is 180° i.e.,
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}=180° \left[\because \angle 1+\angle 2=180°\right] \\ \Rightarrow \angle \mathrm{B}=180°-\angle \mathrm{A}=180°-35° \because \left[\angle \mathrm{A}=35°\right] \\ \Rightarrow \angle \mathrm{B}=145° \end{gathered}$$

Answer:

Given that, the opposite angles of a quadrilateral ABCD are equal. Thus, ABCD is a parallelogram. As per the property of parallelogram, its opposites sides are equal.
$\therefore \mathrm{CD}=\mathrm{AB}=4 \mathrm{cm}$

Hence, CD = 4 cm.

Answer:

Let the quadrilateral be ABCD.
$$\begin{gathered} \angle \mathrm{A}=108°, \\ \angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=x .....\left(1\right) \end{gathered}$$

The sum of angle of quadrilateral is 360°.
$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°$

From (1),
$$\begin{gathered} \Rightarrow 108°+x+x+x=360° \\ \Rightarrow 3x=360°-108°=252° \\ \Rightarrow x=\frac{252°}{3}=84° \end{gathered}$$

Hence, each of the three angles is 84°.

Answer:

Given that, ABCD is a trapezium.
Also, $\mathrm{AB}\parallel \mathrm{DC}$ and $\angle \mathrm{A}=\angle \mathrm{B}=45°$.

As $\mathrm{AB}\parallel \mathrm{DC}$, then BC is transversal.
The sum of two co-interior angle is 180°.
$$\begin{gathered} \therefore \angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{C}=180°-\angle \mathrm{B} \\ \Rightarrow \angle \mathrm{C}=180°-45° \left[\because \angle \mathrm{B}=45°\right] \\ \Rightarrow \angle \mathrm{C}=135° \end{gathered}$$

Similarly, $\angle \mathrm{A}+\angle \mathrm{D}=180°$
$$\begin{gathered} \Rightarrow \angle \mathrm{D}=180°-\angle \mathrm{A}=180°-45° \left[\because \angle \mathrm{A}=45°\right] \\ \Rightarrow \angle \mathrm{D}=135° \end{gathered}$$

Hence, $\angle \mathrm{C} \mathrm{and} \angle \mathrm{D}$ are 135° each.

Answer:

Consider a parallelogram ABCD, in which $\angle \mathrm{ADC} \mathrm{and} \angle \mathrm{ABC}$ are obtuse angles.
Let DE and DF be the two altitudes of parallelogram and angle between them is 60°.

In BEDF quadrilateral,
$\angle \mathrm{BED}=\angle \mathrm{BFD}=90°$.
$\begin{array}{rcl}\therefore \angle \mathrm{FBE}& =& 360°-\left(\angle \mathrm{FDE}+\angle \mathrm{BED}+\angle \mathrm{BFD}\right)\\ & =& 360°-\left(60°+90°+90°\right)\\ & =& 360°-240°\\ & =& 120°\end{array}$

In ABCD parallelogram, opposite angles are equal.
$\angle \mathrm{ADC}=120°$
Now,
$\angle \mathrm{A}+\angle \mathrm{B}=180° \left[\mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{cointerior} \mathrm{angle} \mathrm{is} 180°\right]$
$\begin{array}{rcl}\therefore \angle \mathrm{A}& =& 180°-\angle \mathrm{B}\\ & =& 180°-120° \left[\because \angle \mathrm{FBE}=\angle \mathrm{B}\right]\\ & \Rightarrow & \angle \mathrm{A}=60°\end{array}$

Also,
$\angle \mathrm{C}=\angle \mathrm{A}=60°$

Hence, angle of the parallelogram are 60°, 120°, 60° and 120°.

Answer:

Let the sides of a rhombus be AB = BC = CD = DA = $x$.
Then, join DB.

In $△\mathrm{ALD} \mathrm{and} △\mathrm{BLD}$,
$$\begin{gathered} \angle \mathrm{DLA}=\angle \mathrm{DLB}=90° \left[\because \mathrm{DL} \mathrm{is} \mathrm{the} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{AB}\right] \\ \mathrm{AL}=\mathrm{BL}=\frac{x}{2} \\ \mathrm{and} \mathrm{DL}=\mathrm{DL} \left[\mathrm{Common} \mathrm{side}\right] \\ \therefore △\mathrm{ALD}\cong △\mathrm{BLD} \left[\mathrm{SAS} \mathrm{congruence} \mathrm{rule}\right] \\ \Rightarrow \mathrm{AD}=\mathrm{BD} \left[\mathrm{by} \mathrm{CPCT}\right] \end{gathered}$$

$\mathrm{In} △\mathrm{ADB}, \mathrm{AD}=\mathrm{AB}=\mathrm{DB}=x$.
Then, $△\mathrm{ADB}$ is an equilateral triangle
$\therefore \angle \mathrm{A}=\angle \mathrm{ADB}=\angle \mathrm{ABD}=60°$

Similarly, $△$DBC is an equilateral triangle.
$\therefore \angle \mathrm{C}=\angle \mathrm{BDC}=\angle \mathrm{DBC}=60°$
Also, $\angle \mathrm{A}=\angle \mathrm{C}$.
$\begin{array}{rcl}\therefore \angle \mathrm{D}& =& \angle \mathrm{B}\\ & =& 180°-60° \left[\because \mathrm{sum} \mathrm{of} \mathrm{interior} \mathrm{angles} \mathrm{is} 180°\right]\\ & =& 120°\end{array}$

Hence, $\angle \mathrm{A}=60°, \angle \mathrm{B}=120°, \angle \mathrm{C}=60° \mathrm{and} \angle \mathrm{D}=120°$.
 

Answer:

Given that, ABCD is a parallelogram and AE = CF.
To show: OE = OF

Join BD, meet AC at point O.
Since, the diagonals of a parallelogram bisect each other.

$\therefore \mathrm{OA}=\mathrm{OC} \mathrm{and} \mathrm{OD}=\mathrm{OB}$
Now, OA = OC and AE = CF
$$\begin{gathered} \Rightarrow \mathrm{OA}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF} \\ \Rightarrow \mathrm{OE}=\mathrm{OF} \end{gathered}$$
Thus, BEDE is a quadrilateral whose diagonals bisect each other.
Therefore, BFDE is a parallelogram.
Hence, proved.

Answer:

Given that, ABCD is a trapezium
such that $\mathrm{AB}\parallel \mathrm{DC} \mathrm{and} \mathrm{EF}\parallel \mathrm{AB}\parallel \mathrm{CD}$.
Join, the diagonal AC which intersects EF at O.

In $△$ADC,
E is mid-point of AD and $\mathrm{OE}\parallel \mathrm{CD}$.

By mid-point theorem, O is the mid point of AC.

In $△$CBA,
O is mid point of AC and $\mathrm{OF}\parallel \mathrm{AB}.$

So, by mid-point theorem, E is the mid-point of BC.

Hence, proved.

Answer:

In $△$ABC,
$\mathrm{PQ}\parallel \mathrm{AB},\mathrm{PR}\parallel \mathrm{AC} \mathrm{and} \mathrm{RQ}\parallel \mathrm{BC}$.

In quadrilateral BCAR, BR ∥ CA and BC ∥ RA.
Then, BCAR is a parallelogram.

∴ BC = AR           .....(1)

So, in quadrilateral BCQA, BC ∥ AQ and AB ∥ QC.

Thus, BCQA is a parallelogram.
∴ BC = AQ          .....(2)

On adding (1) and (2),
2BC = AR + AQ
$$\begin{gathered} \Rightarrow 2\mathrm{BC}=\mathrm{RQ} \\ \Rightarrow \mathrm{BC}=\frac{1}{2}\mathrm{QR} \end{gathered}$$
Hence, proved

Answer:

In equilateral $△$ABC, D, E and F are the mid-point of sides BC, CA and AB respectively.

In $△$ABC, E and F are the mid-point of AC and AB.
Then, using mid point theorem,
EF ∥ BC and EF = $\frac{1}{2}\mathrm{BC} .....\left(1\right)$

Similarly,
DF$\parallel$AC, DE$\parallel$AB
$\mathrm{DE}=\frac{1}{2}\mathrm{AB} \mathrm{and} \mathrm{FD}=\frac{1}{2}\mathrm{AC}$          .....(2)

As, $△$ABC is an equilateral triangle.
∴ AB = BC = CA
$\Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{CA}$      [dividing by 2]
$\Rightarrow \mathrm{DE}=\mathrm{EF}=\mathrm{FD}$            [from (1) and (2)]
Thus, all sides of AEDF are equal.
Therefore, $△$DEF is an equilateral triangle.
Hence, proved.

Answer:

Given that, ABCD is a parallelogram and AP = CQ.

In $△$AMP and $△$CMQ,
$\angle \mathrm{MAP}=\angle \mathrm{MCQ}$ [alternate interior angles]
AP = CQ            (given)
and $\angle \mathrm{AMP}=\angle \mathrm{CMQ}$  [vertically opposite angles]
$\therefore △\mathrm{AMP}\cong △\mathrm{CMQ}$  [AAS congruence rule]

$\Rightarrow$AM = CM                    [CPCT rule]
and PM = MQ                 [CPCT rule]
Thus, AC and PQ bisect each other.
Hence, proved.

Answer:

Given that ABCD is a parallelogram, P is a mid-point of BC such that $\angle \mathrm{BAP}=\angle \mathrm{DAP}$.
Since, ABCD is a parallelogram.
$\therefore \mathrm{AD}\parallel \mathrm{BC}$ and AB is transversal, then
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}=180° \left[\mathrm{sum} \mathrm{of} \mathrm{cointerior} \mathrm{angles} \mathrm{is} 180°\right] \\ \Rightarrow \angle \mathrm{B}=180°-\angle \mathrm{A} .....\left(1\right) \end{gathered}$$

In $△$ABP,
$$\begin{gathered} \angle \mathrm{PAB}+\angle \mathrm{B}+\angle \mathrm{BPA}=180° \left[\mathrm{by} \mathrm{angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangle}\right] \\ \Rightarrow \frac{1}{2}\angle \mathrm{A}+180°-\angle \mathrm{A}+\angle \mathrm{BPA}=180° \left[\mathrm{From} \left(1\right)\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow \angle \mathrm{BPA}-\frac{\angle \mathrm{A}}{2}=0° \\ \Rightarrow \angle \mathrm{BPA}=\frac{\angle \mathrm{A}}{2} .....\left(2\right) \\ \Rightarrow \angle \mathrm{BPA}=\angle \mathrm{BAP} \\ \Rightarrow \mathrm{AB}=\mathrm{BP} [\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{equal} \mathrm{angles} \mathrm{are} \mathrm{equal}] \end{gathered}$$

On multiplying both sides by 2, we get
2AB = 2BP
$$\begin{gathered} \Rightarrow 2\mathrm{AB}=\mathrm{BC} \left[\mathrm{As} \mathrm{P} \mathrm{is} \mathrm{the} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{BC}\right] \\ \Rightarrow 2\mathrm{CD}=\mathrm{AD} \end{gathered}$$
[since, ABCD is a parallelogram, then AB = CD and BC = AD]
Hence, proved.

Answer:

In an isosceles triangle $△$ABC, a square ADEF is inscribed.
To prove: CE = BE
Proof: In the isosceles $△$ABC, $\angle \mathrm{A}=90°$
and AB = AC     .....(1)

Since, ADEF is a square.
AD = AF          [all sides of squares are equal]     .....(2)

On subtracting (2) from (1),
AB − AD = AC − AE 
⇒ BD = CF               .....(3)

Now, in $△\mathrm{CFE} \mathrm{and} △\mathrm{BDE}$,

BD = CF         [From (3)]
DE = EF         [sides of square]
and $\angle \mathrm{CFE}=\angle \mathrm{EDB}=90°$
$\therefore △\mathrm{CFE}\cong △\mathrm{BDE} \left[\mathrm{by} \mathrm{SAS} \mathrm{congruence} \mathrm{rule}\right]$
$\Rightarrow \mathrm{CE}=\mathrm{BE} \left[\mathrm{by} \mathrm{CPCT}\right]$
Hence, vertex E of the square bisects the hypotenuse BC.

Answer:

Given, a parallelogram ABCD in which AB = 10 cm and AD = 6 cm.
Now, draw a bisector $\angle \mathrm{A}$ meets DC in E and produces it to F and produce BC to meet at F.

Also, produce AD to H and join HF, so that ABFH, is a parallelogram.
Since, $\mathrm{HF}\parallel \mathrm{AB}$.
$$\begin{gathered} \angle \mathrm{AFH}=\angle \mathrm{FAB} \left[\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right] \\ \angle \mathrm{HAF}=\angle \mathrm{FAB} \left[\mathrm{AF} \mathrm{is} \mathrm{bisector} \mathrm{of} \angle \mathrm{A}\right] \\ \Rightarrow \angle \mathrm{HAF}=\angle \mathrm{AFH} \\ \Rightarrow \mathrm{HF}=\mathrm{AH} \left[\because \mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles} \mathrm{are} \mathrm{equal}\right] \end{gathered}$$

But
$$\begin{gathered} \mathrm{HF}=\mathrm{AB}=10 \mathrm{cm} \\ \mathrm{AH}=\mathrm{HF}=10 \mathrm{cm} \\ \mathrm{AD}+\mathrm{DH}=10 \mathrm{cm} \\ \Rightarrow \mathrm{DH}=(10 - 6) \mathrm{cm}=4 \mathrm{cm} \end{gathered}$$

Since, CFHD is a parallelogram.
Thus, the opposite side are equal.
Hence, DH = CF = 4 cm

Answer:

Let ABCD be the quadrilateral, P, Q, R and S are the mid-point of sides AB, BC, CD and DA. Also, AC = BD.

Proof : In $△$ADC, S and R are the mid-points of AD and DC respectively.
Then by mid point theorem,
$\mathrm{SR}\parallel \mathrm{AC} \mathrm{and} \mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(1\right)$

In $△$ABC, P and Q are mid-point of AB and BC respectively.
Then by mid-point theorem,
$\mathrm{PQ}\parallel \mathrm{AC} \mathrm{and} \mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(2\right)$

From (1) and (2),
SR = PQ = $\frac{1}{2}\mathrm{AC} .....\left(3\right)$

Similarly, in $△$BCD,
$\mathrm{RQ}\parallel \mathrm{BD} \mathrm{and} \mathrm{RQ}=\frac{1}{2}\mathrm{BD} .....\left(4\right)$
and in $△$BAD,
$\mathrm{SP}\parallel \mathrm{BD} \mathrm{and} \mathrm{SP}=\frac{1}{2}\mathrm{BD} .....\left(5\right)$

From (4) and (5),
$\mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{BD}=\frac{1}{2}\mathrm{AC} \left[\mathrm{given}, \mathrm{AC}=\mathrm{BD}\right] .....\left(6\right)$

From (3) and (6),
$\mathrm{SR}=\mathrm{PQ}=\mathrm{SP}=\mathrm{RQ}$
Thus, all the sides of a quadrilateral PQRS are equal.
Hence, PQRS is a rhombus.

Answer:

Given: A quadrilateral ABCD with P, Q, R and S as the mid-points of sides AB, BC, CD and DA respectively.
Also, $\mathrm{AC}\perp \mathrm{BD}$.
To prove: PQRS is a rectangle 
Proof: As $\mathrm{AC}\perp \mathrm{BD}$
$\therefore \angle \mathrm{COD}=\angle \mathrm{AOD}=\angle \mathrm{AOB}=\angle \mathrm{COB}=90°$

In $△$ADC, S and R are mid-points of AD and CD respectively. Then by mid-point theorem,
$\mathrm{SR}\parallel \mathrm{AC} \mathrm{and} \mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(1\right)$

In $△$ABC, P and Q are mid-points of AB and BC respectively. Then by mid-point theorem,
$\mathrm{PQ}\parallel \mathrm{AC} \mathrm{and} \mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(2\right)$

From (1) and (2),
$\mathrm{PQ}\parallel \mathrm{SR} \mathrm{and} \mathrm{PQ}=\mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(3\right)$

Similarly,
$\mathrm{SP}\parallel \mathrm{RQ} \mathrm{and} \mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{BD} .....\left(4\right)$

Then, in quadrilateral PQRS,
$\mathrm{PQ}\parallel \mathrm{RS} \mathrm{and} \mathrm{PS}\parallel \mathrm{RQ}$

Since, AC ∥ PQ.
$\therefore \angle \mathrm{P}=\angle \mathrm{Q}=90°$
So, PQRS is a rectangle.
Hence, proved.

Answer:

Given: A quadrilateral ABCD with P, Q, R and S as the mid-points of sides AB, BC, CD and DA respectively.
Also, AC = BD and AC$\perp$BD.
To prove: PQRS is a square.
Proof: In $△$ADC, S and R are the mid-points of AD and CD respectively.
Then by mid-point theorem,
$\mathrm{SR}\parallel \mathrm{AC} \mathrm{and} \mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(1\right)$

In $△$ABC, P and Q are the mid-points of AB and BC respectively.
Then by mid-point theorem, 
$\mathrm{PQ}\parallel \mathrm{AC} \mathrm{and} \mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(2\right)$

From (1) and (2),
$\mathrm{PQ}\parallel \mathrm{SR} \mathrm{and} \mathrm{PQ}=\mathrm{SR}=\frac{1}{2}\mathrm{AC} .....\left(3\right)$

Similarly in $△$ABD, by mid point theorem,
$\mathrm{SP}\parallel \mathrm{BD} \mathrm{and} \mathrm{SP}=\frac{1}{2}\mathrm{BD}=\frac{1}{2}\mathrm{AC} \left[\because \mathrm{AC}=\mathrm{BD}\right] .....\left(4\right)$

In $△$BCD, by mid point theorem,
$\mathrm{RQ}\parallel \mathrm{BD} \mathrm{and} \mathrm{RQ}=\frac{1}{2}\mathrm{BD}=\frac{1}{2}\mathrm{AC} \left[\because \mathrm{BD}=\mathrm{AC}\right] .....\left(5\right)$

From (4) and (5),
$\mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{AC} .....\left(6\right)$

From (3) and (6),
PQ = SR = SP = RQ
Thus, all 4 sides are equal.

Now, in quadrilateral OERF
$\mathrm{PQ}\parallel \mathrm{RS} \mathrm{and} \mathrm{RQ}\parallel \mathrm{PS}$
Since, AC ∥ PQ.
∴ ∠P = ∠Q = 90°

So, PQRS is a square.
Hence, proved.

Answer:

Consider a parallelogram ABCD such that its diagonal AC bisects ∠A.
Thus, $\angle \mathrm{CAB}=\angle \mathrm{CAD} .....\left(1\right)$

To show: ABCD is a rhombus.
Proof: Since, ABCD is a parallelogram.
Therefore, AB ∥ CD and AC is the transversal.
$\therefore \angle \mathrm{CAB}=\angle \mathrm{ACD} \left[\mathrm{alternate} \mathrm{interior} \mathrm{angles}\right]$

Again, AD ∥ BC and AC is the transversal.
$\therefore \angle \mathrm{CAD}=\angle \mathrm{ACB} \left[\mathrm{alternate} \mathrm{interior} \mathrm{angles}\right]$

So,
$\angle \mathrm{ACD}=\angle \mathrm{ACB} \left[\because \angle \mathrm{CAB}=\angle \mathrm{CAD}\right] .....\left(2\right)$

Also,
$\angle \mathrm{A}=\angle \mathrm{C} \left[\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}\right]$
$$\begin{gathered} \Rightarrow \frac{1}{2}\angle \mathrm{A}=\frac{1}{2}\angle \mathrm{C} \\ \Rightarrow \angle \mathrm{DAC}=\angle \mathrm{DCA} \left[\mathrm{from} \left(1\right) \mathrm{and} \left(2\right)\right] \\ \Rightarrow \mathrm{CD}=\mathrm{AD} \left[\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{the} \mathrm{equal} \mathrm{angles} \mathrm{are} \mathrm{equal}\right] \end{gathered}$$

As, AB = CD and AD = BC [opposite side of parallelogram are equal]
$\therefore$ AB = BC = CD = AD

Thus, all sides are equal.
So, ABCD is a rhombus.
Hence, proved.

Answer:

Given that, a parallelogram ABCD such that P and Q are the mid-points of AB and CD respectively.
To show: PQRS is a parallelogram.
Proof: As, ABCD is a parallelogram.
⇒ AB$\parallel$CD
⇒ AP$\parallel$QC
Also, AB = DC as the opposite sides of a parallelogram are equal.

$$\begin{gathered} \Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{DC} \\ \Rightarrow \mathrm{AP}=\mathrm{QC} \left[\because \mathrm{P} \mathrm{and} \mathrm{Q} \mathrm{are} \mathrm{the} \mathrm{midpoints} \mathrm{of} \mathrm{AB} \mathrm{and} \mathrm{DC} \mathrm{respectively}\right] \end{gathered}$$
Here, AP ∥ QC and AP = QC.
Thus, APCQ is a parallelogram.


∴ AQ ∥ PC or SQ ∥ PR          .....(1)

Again, AB ∥ DC or BP ∥ DQ.

Also, AB = DC.
$\Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{DC}$          [dividing both sides by 2]
⇒ BP = QD               [since, P and Q are the mid-point of AB and DC]
Here, BP ∥ QD and BP = QD.
So, BPDQ is a parallelogram.
$\therefore \mathrm{PD}\parallel \mathrm{BQ} \mathrm{or} \mathrm{PS}\parallel \mathrm{QR} .....\left(2\right)$

From (1) and (2), PRQS is a parallelogram.
Hence, proved.
 

Answer:

Given that, ABCD is a quadrilateral with AB ∥ DC and AD = BC.
Construction: Extend AB to E and draw a line CE parallel to AD.

Proof: As, AD ∥ CE and transversal AE cuts them at A and E.
Thus,
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{E}=180° \left[\because \mathrm{sum} \mathrm{of} \mathrm{cointerior} \mathrm{angles} \mathrm{is} 180°\right] \\ \Rightarrow \angle \mathrm{A}=180°-\angle \mathrm{E} .....\left(1\right) \end{gathered}$$

Now, AB ∥ CD and AD ∥ CE.
So, quadrilateral AECD is a parallelogram.

$$\begin{gathered} \Rightarrow \mathrm{AD}=\mathrm{CE} \\ \Rightarrow \mathrm{BC}=\mathrm{CE} \left[\because \mathrm{AD}=\mathrm{BC}\right] \end{gathered}$$

In $△$BCE,
CE = BC                     [shown above]
$\Rightarrow \angle \mathrm{CBE}=\angle \mathrm{CEB}$       [opposite angles of equal sides are equal]
$$\begin{gathered} \Rightarrow 180°-\angle \mathrm{ABC}=\angle \mathrm{E} \left[\because \angle \mathrm{ABC}+\angle \mathrm{CBE}=180°\right] \\ \Rightarrow 180°-\angle \mathrm{E}=\angle \mathrm{B}=\angle \mathrm{ABC} .....\left(2\right) \end{gathered}$$

From (1) and (2),
$\angle \mathrm{A}=\angle \mathrm{B}$
Hence, proved.

Answer:

In the given figure,
$$\begin{gathered} \mathrm{AB}\parallel \mathrm{DE} \mathrm{and} \mathrm{AC}\parallel \mathrm{DF} \\ \mathrm{AB}=\mathrm{DE} \mathrm{and} \mathrm{AC}=\mathrm{DF} \end{gathered}$$
To prove: $\mathrm{BC}\parallel \mathrm{EF} \mathrm{and} \mathrm{BC}=\mathrm{EF}$
Proof: In quadrilateral ABED,
$\mathrm{AB}\parallel \mathrm{DE} \mathrm{and} \mathrm{AB}=\mathrm{DE}$
Thus, ABED is a parallelogram. 
$\Rightarrow \mathrm{AD}\parallel \mathrm{BE} \mathrm{and} \mathrm{AD}=\mathrm{BE}$

Now, in quadrilateral ACFD,
$\mathrm{AC}\parallel \mathrm{FD} \mathrm{and} \mathrm{AC}=\mathrm{FD} .....\left(1\right)$
Thus, ACFD is a parallelogram.
$\Rightarrow \mathrm{AD}\parallel \mathrm{CF} \mathrm{and} \mathrm{AD}=\mathrm{CF} .....\left(2\right)$

From (1) and (2),
$\mathrm{AD}=\mathrm{BE}=\mathrm{CF} \mathrm{and} \mathrm{CF}\parallel \mathrm{BE} .....\left(3\right)$

As, in quadrilateral BCFE,
BE = CF and BE$\parallel$CF           [from (3)]
Thus, BCFE is a parallelogram.
Hence, proved.

Answer:

In $△$ABC, AD is a median and E is the mid-point of AD.
Construction: Draw DP ∥ EF.

Proof: In $△$ADP,
E is the mid-point of AD and EF ∥ DP.
Thus, by converse of mid-point theorem, F is mid-point of AP.

In $△$FBC,
D is mid-point of BC and DP ∥ BF.
So, by converse of mid-point theorem, P is mid-point of FC.
Thus, AF = FP = PC.
$\Rightarrow \mathrm{AF}=\frac{1}{3}\mathrm{AC}$
Hence, proved.

Answer:

In the square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
To show: PQRS is a square.
Construction : Join AC and BD.


Proof : As, ABCD is a square.
$\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$

Also, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.

Now, in $△$ADC, by mid-point theorem,
SR ∥ AC
and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$      .....(1)

In $∆\mathrm{ABC}, \mathrm{PQ}\parallel \mathrm{AC}$$△$ABC,
PQ ∥ AC
and $\mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(2\right)$

From (1) and (2),
$\mathrm{SR}\parallel \mathrm{PQ} \mathrm{and} \mathrm{SR}=\mathrm{PQ}=\frac{1}{2}\mathrm{AC} .....\left(3\right)$

Similarly, $\mathrm{SP}\parallel \mathrm{BD} \mathrm{and} \mathrm{BD}\parallel \mathrm{RQ}$
$\therefore \mathrm{SP}\parallel \mathrm{RQ} \mathrm{and} \mathrm{SP}=\frac{1}{2}\mathrm{BD}$
and $\mathrm{RQ}=\frac{1}{2}\mathrm{BD}$
$\therefore \mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{BD}$

Thus, the diagonals of a square bisect each other at 90° and they are equal.
$$\begin{gathered} \therefore \mathrm{AC}=\mathrm{BD} \\ \Rightarrow \mathrm{SP}=\mathrm{RQ}=\frac{1}{2}\mathrm{AC} .....\left(4\right) \end{gathered}$$

From (3) and (4),
$\mathrm{SR}=\mathrm{PQ}=\mathrm{SP}=\mathrm{RQ} \left[\mathrm{all} \mathrm{side}s \mathrm{are} \mathrm{equal}\right]$

Now, in quadrilateral OERF, 
$\mathrm{OE}\parallel \mathrm{FR} \mathrm{and} \mathrm{OF}\parallel \mathrm{ER}$
$\therefore \angle \mathrm{EOF}=\angle \mathrm{ERF}=90°$
Thus, PQRS is a square.
Hence, proved.

Answer:

Given: ABCD is a trapezium in which AB ∥ CD.
Also, E and F are the mid-points of sides AD and BC respectively.
To prove: EF ∥ AB and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)$
Construction: Join BE and extend it to meet CD produced at G. Thus draw BD with intersects EF at O.

Proof: In $△$GCB, E and F are the mid-points of BG and BC respectively.
Then by mid-point theorem,
$\mathrm{EF}\parallel \mathrm{GC}$
But $\mathrm{GC}\parallel \mathrm{AB} \mathrm{or} \mathrm{CD}\parallel \mathrm{AB} \left[\mathrm{given}\right]$
$\therefore \mathrm{EF}\parallel \mathrm{AB}$

In $△$ADB,
$\mathrm{AB}\parallel \mathrm{EO}$ and E is the mid-point of AD.
Hence, by converse of mid-point theorem, O is mid-point of BD.
Also, $\mathrm{EQ}=\frac{1}{2}\mathrm{AB} .....\left(1\right)$ 

In $△$BDC,
OF ∥ CD and O is the mid-point of BD.
$\therefore \mathrm{OF}=\frac{1}{2}\mathrm{CD}$ [by converse of mid-point theorem]      ......(2)

On adding (1) and (2),
$\mathrm{EO}+\mathrm{OF}=\frac{1}{2}\mathrm{AB}+\frac{1}{2}\mathrm{CD}$
$\Rightarrow \mathrm{EF}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)$
Hence, proved.
 

Answer:

Let ABCD be a parallelogram and AP, BR, CR and DP be are the bisectors $\angle \mathrm{A},\angle \mathrm{B},\angle \mathrm{C} \mathrm{and} \angle \mathrm{D}$.
To prove: Quadrilateral PQRS is a rectangle.


Proof: As, ABCD is a  parallelogram. Therefore, DC$\parallel$AB and DA is the transversal.
$\therefore \angle \mathrm{A}+\angle \mathrm{D}=180°$            [sum of co-interior angles of a parallelogram is 180°]
$$\begin{gathered} \Rightarrow \frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{D}=90° \\ \therefore \angle \mathrm{PAD}+\angle \mathrm{PDA}=90° \\ \Rightarrow \angle \mathrm{APD}=90° \left[\mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{is} 180°\right] \\ \therefore \angle \mathrm{SPQ}=90° \left[\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right] \\ \Rightarrow \angle \mathrm{PQR}=90° \\ \Rightarrow \angle \mathrm{QRS}=90° \\ \mathrm{and} \angle \mathrm{PSR}=90° \end{gathered}$$
Thus, PQRS is a quadrilateral with each angle is 90°, i.e., right angle.
Hence, PQRS is a rectangle.

Answer:

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To show: PQ is bisected at O.

In $△\mathrm{ODP} \mathrm{and} △\mathrm{OBQ}$,
$\angle \mathrm{BOQ}=\angle \mathrm{POD} \left[\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right]$ 
$\angle \mathrm{OBQ}=\angle \mathrm{ODP} \left[\mathrm{alternate} \mathrm{interior} \mathrm{angles}\right]$
and OB = OD        [∵ diagonals of a parallelogarm bisect each other]
$\therefore △\mathrm{ODP}\cong △\mathrm{OBQ} \left[\mathrm{ASA} \mathrm{congruence} \mathrm{rule}\right]$
$\Rightarrow \mathrm{OP}=\mathrm{OQ} \left[\mathrm{CPCT} \mathrm{rule}\right]$
Hence, proved.

 

Answer:

Given: Rectangle ABCD, diagonal BD bisects $\angle \mathrm{B}$.
To show: ABCD is a square.
Construction: Join AC.

Proof: In $△\mathrm{BAD} \mathrm{and} △\mathrm{BCD},$
$\angle \mathrm{A}=\angle \mathrm{C}=90°$
BD = BD          (common)
AB = CD          (opposite sides of a rectangle are equal)
$\therefore △\mathrm{BAD}\cong △\mathrm{BCD} \left[\mathrm{RHS} \mathrm{congruence} \mathrm{rule}\right]$
Thus, AB = BC
and AD = CD            [by CPCT rule]

Therefore,
AB = BC = CD = DA.
So, ABCD is a square.
Hence, proved.
 

Answer:

In $△$ABC, D, E, F are the mid-points of the sides AB, BC and CA respectively.
To prove: $△$ABC is divided into four congruent triangles.

Proof: As, $△$ABC is a triangle and D, E and F are the mid-point of sides AB, BC and CA respectively.
Then,
AD = BD = $\frac{1}{2}$AB,
BE = EC = $\frac{1}{2}$BC
and AF = CF = $\frac{1}{2}$AC

Then, using the mid-point theorem,
$$\begin{gathered} \mathrm{EF}\parallel \mathrm{AB} \mathrm{and} \mathrm{EF}=\frac{1}{2}\mathrm{AB}=\mathrm{AD}=\mathrm{BD} \\ \mathrm{ED}\parallel \mathrm{AC} \mathrm{and} \mathrm{ED}=\frac{1}{2}\mathrm{AC}=\mathrm{AF}=\mathrm{CF} \end{gathered}$$
and $\mathrm{DF}\parallel \mathrm{BC} \mathrm{and} \mathrm{DF}=\frac{1}{2}\mathrm{BC}=\mathrm{BE}=\mathrm{CE}$

In $△$ADF and $△$EFD,
AD = EF
AF = DE
DF = DF  (common)
$$\begin{gathered} \therefore △\mathrm{ADF}\cong △\mathrm{EFD} \left[\mathrm{SSS} \mathrm{congruence} \mathrm{rule}\right] \\ \mathrm{Similrly},△\mathrm{DEF}\cong △\mathrm{EDB} \\ \mathrm{and} △\mathrm{DEF}\cong △\mathrm{CFE} \end{gathered}$$

So, $△$ABC is divided into four congruent triangles.
Hence, proved.
 

Answer:

Consider a trapezium ABCD in which AB ∥ DC and let M and N be the mid-points of the diagonals AC and BD respectively.
To prove: MN ∥ AB ∥ CD
Construction: Join CN and extend it to meet AB at E as shown:

Proof: In $△\mathrm{CDN} \mathrm{and} △\mathrm{EBN}$,
$\mathrm{DN}=\mathrm{BN}$        [∵ N is the mid-point of BD]
$\angle \mathrm{DCN}=\angle \mathrm{BEN}$       [alternate interior angles]
and $\angle \mathrm{CDN}=\angle \mathrm{EBN}$        [alternate interior angles]
$\therefore △\mathrm{CDN}\cong △\mathrm{EBN}$        [AAS congruence rule]
$\Rightarrow \mathrm{DC}= \mathrm{EB} \mathrm{and} \mathrm{CN}=\mathrm{NE}$       [by CPCT rule]

Thus, in $△\mathrm{CAE}$, the points M and N are the mid-points of AC and CE respectively.
$\therefore \mathrm{MN}\parallel \mathrm{AE}$           [by mid-point theorem]
$\Rightarrow$MN$\parallel$AB$\parallel$CD
Hence, proved.

Answer:

Given: A parallelogram ABCD where P is the mid-point of DC.
To prove: DA = AR and CQ = QR 
Proof: ABCD is a parallelogram.
$\therefore \mathrm{BC}=\mathrm{AD} \mathrm{and} \mathrm{BC}\parallel \mathrm{AD}$

Also, $\mathrm{DC}=\mathrm{AB} \mathrm{and} \mathrm{DC}\parallel \mathrm{AB}$

As, P is the mid-point of DC.
$\therefore \mathrm{DP}=\mathrm{PC}=\frac{1}{2}\mathrm{DC}$

Now, $\mathrm{QC}\parallel \mathrm{AP} \mathrm{and} \mathrm{PC}\parallel \mathrm{AQ}$.
Thus, APCQ is a parallelogram.

Therefore, $\mathrm{AQ}=\mathrm{PC}=\frac{1}{2}\mathrm{DC}=\frac{1}{2}\mathrm{AB}=\mathrm{BQ} \left[\because \mathrm{DC}=\mathrm{AB}\right] .....\left(1\right)$

Now, in $△\mathrm{AQR} \mathrm{and} △\mathrm{BQC}$,
$\mathrm{AQ}=\mathrm{BQ} \left[\mathrm{From} \left(1\right)\right]$
$\angle \mathrm{AQR}=\angle \mathrm{BQC} \left[\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right]$
and $\angle \mathrm{ARQ}=\angle \mathrm{BCQ} \left[\mathrm{alternate} \mathrm{interior} \mathrm{angles}\right]$
$\therefore △\mathrm{AQR}\cong △\mathrm{BQC} \left[\mathrm{AAS} \mathrm{congruence} \mathrm{rule}\right]$
Then, AR = BC  [by CPCT rule]
But BC = DA
$\therefore$ AR = DA
and CQ = QR  [CPCT rule]
Hence, proved.