Rd Sharma 2022 for Class 9 Maths Chapter 12 - Congruent Triangles

Answer:

In, it is given that

, and

We have to find, and

Since and

Then (as AB = AC)

Now 

(By property of triangle)

Thus, 

, as (given)

So,

Since,, so

Hence .

Answer:

In it is given that

, and

We have to find.

Since

Then (isosceles triangles)

Now 

(As given)

Thus 

(Property of triangle)

Hence .

Answer:

It is given that

We have to find.

(Isosceles triangle)

Now 

Since exterior angle of isosceles triangle is the sum of two internal base angles

Now 

So, (By property of triangle)

Hence .

Answer:

We have to find the measure of each exterior angle of an equilateral triangle.

It is given that the triangle is equilateral

So, and 

Since triangle is equilateral 

So,

Now we have to find the exterior angle.

As we know that exterior angle of the triangle is sum of two interior angles

Thus

Hence each exterior angle is.

Answer:

It is given that the base of an isosceles triangle is produced on both sides.

We have to prove that the exterior angles so formed are equal to each other.

That is we need to show that

Let the is isosceles having base and equal sides AB and AC

Then, and

(Isosceles triangles)

Now 

    .........(1)

And, 

   .......(2)

Thus 

  ........(3)

Now from equation (2)

     .........(4)

Since

Hence from equation (3) and (4)

Answer:

It is given that

And is the mid point of

So

And is the mid point of

So

And is the mid point of

So

We have to prove that

In we have 

(Equilateral triangle)

Then 

, and

, and

Similarly comparing and we have

, and

And (Since N is the mid point of )

So by congruence criterion, we have 

Hence.

Answer:

We have to prove that the median of an equilateral triangle are equal.

Let be an equilateral triangle with as its medians.

Let

In we have

(Since similarly)

(In equilateral triangle, each angle)

And (common side)

So by congruence criterion we have

This implies that, 

Similarly we have

Hence .

Answer:

It is given that

We have to find the ratio.

Since

And

So we have,

So

Hence .

Answer:

It is given that

Is right angled triangle

And 

We have to find and

Since

(Isosceles triangle)

Now 

(Property of triangle)

()

So

Hence 

Answer:

It is given that

Δis a square and Δ is an equilateral triangle.

We have to prove that

(1) and (2)

(1)

Since,

(Angle of square)

(Angle of equilateral triangle)

Now, adding both

Similarly, we have

Thus in and we have

(Side of square)

And (equilateral triangle side)

So by congruence criterion we have

Hence.

(2)
Since 
QR = RS ( Sides of Square)
RS = RT (Sides of Equilateral triangle)

We get
QR = RT

Thus, we get
$\angle TQR=\angle RTQ$  (Angles opposite to equal sides are equal)

Now, in the triangle TQR, we have

$$\begin{gathered} \angle TQR+\angle RTQ+\angle QRT={180}^0 \\ \angle TQR+\angle TQR+{150}^0={180}^0 \\ 2\angle TQR+{150}^0={180}^0 \\ 2\angle TQR={180}^0-{150}^0 \\ 2\angle TQR={30}^0 \\ \angle TQR=\frac{{30}^0}{2}={15}^0 \end{gathered}$$

Answer:

It is given that

P and Q are equidistant from A and B that is

, and

We are asked to show that line PO is perpendicular bisector of line AB.

First of all we will show that ΔAQP and ΔQBP are congruent to each other and ultimately we get the result.

Consider the triangles AQP and QBP in which

AP=BP, AQ=BQ, PQ=PQ

So by SSS property we have

Implies that

Now consider the triangles ΔAPC and ΔPCB in which

And

So by SAS criterion we find that,

So this implies that AC=BC and

But 

Hence PQ is perpendicular bisector of AB.

Answer:

It is given that 

Is bisector of and is bisector of.

And is isosceles with

We have to prove that

If will be sufficient to prove to show that

Now in these two triangles

Since, so

Now as BD and CE are bisector of the respectively, so

, and

BC=BC

So by congruence criterion we have 

Hence Proved.

Answer:

It is given that

We have to prove that the lines and bisect at.

If we prove that, then

We can prove and bisects at.

Now in and

(Given)

(Since and is transversal)

And (since and is transversal)

So by congruence criterion we have,

, so

Hence and bisect each other at.

Answer:

It is given that 

We have to prove that

Now

In we have

(Isosceles triangle)     .......(1)

Now we have

(Vertically opposite angles)

(Since, given)

    .......(2)

Subtracting equation (2) from equation (1) we have 

Now in and we have

(Given)

So all the criterion for the two triangles and are satisfied to be congruent

Hence by congruence criterion we have proved.

Answer:

It is given that 

We are asked to show that

Let us assume

, and are right angled triangle.

Thus in and, we have

And (given)

Hence by AAs congruence criterion we haveProved.

Answer:

We have to prove each angle of an equilateral triangle is.

Here 

(Side of equilateral triangle)

      ...........(1)

And 

(Side of equilateral triangle)

         ..........(2)

From equation (1) and (2) we have 

Hence

Now

That is (since)

Hence Proved.

Answer:

It is given that 

We have to prove that triangle ΔABC is equilateral.

Since (Given)

So,     ..........(1)

And (given)

So       ........(2)

From equation (1) and (2) we have 

Now from above equation if we have

Given condition satisfy the criteria of equilateral triangle.

Hence the given triangle is equilateral.

Answer:

It is given that 

We have to find and.

Since so,

Now (property of triangle)

(Since )

Here

Then

Hence 

Answer:

We have to prove that is isosceles.

Let Δ be such that the bisector of is parallel to

The base, we have 

(Corresponding angles)

(Alternate angle)

(Since)

Hence is isosceles.

Answer:

In the triangle ABC it is given that the vertex angle is twice of base angle.

We have to calculate the angles of triangle.

Now, let be an isosceles triangle such that

Then 

(Given)

() 

Now (property of triangle)

Hence 

Answer:

It is given that

We have to prove

In we have 

(Given)

So,

Now (Given)

Since corresponding angle are equal, so

That is, 

Henceproved.

Answer:

It is given that

In,

We have to prove that

Now 

(Given)

Thus

   ........(1)

In, we have

So, {from equation (1)}

Hence Proved.

Answer:

In the following figure it is given that sides AB and PQ are parallel and BP is bisector of

We have to prove that is an isosceles triangle.

(Since BP is the bisector of)    ........(1)

(Since and are parallel)     .......(2)

Now from equation (1) and (2) we have

So

Now since and is a side of.

And since two sides and are equal, so

Hence is an isosceles triangle.

Answer:

It is given that in

And bisects

We have to prove that

Now let

(Given)

Sinceis a bisector of so let

Let be the bisector of

If we join we have 

In

So

In triangle and we have

(Given)

(Proved above)

So by congruence criterion, we have

And

, and (since)

In we have 

Since,

And,

So,

In we have 

Here, 

HenceProved.

Answer:

∠ABD is the external angle adjacent to ∠ABC.

∆ABC is an isosceles triangle.

AB = AC               (Given)

∴ ∠C = ∠ABC      .....(1)        (In a triangle, equal sides have equal angles opposite to them)

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{ABC}}{2} .....\left(2\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2} .....\left(3\right)$

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º               (Angle sum property of triangle)

$$\begin{gathered} \Rightarrow \frac{\angle \mathrm{ABC}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(2\right) \mathrm{and} \left(3\right)\right] \\ \Rightarrow \frac{\angle \mathrm{ABC}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \frac{2\angle \mathrm{ABC}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(1\right)\right] \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BOC}=180° .....\left(4\right) \end{gathered}$$

Now,

∠ABD + ∠ABC = 180º             .....(5)        (Linear pair)

From (4) and (5), we have

∠ABD + ∠ABC = ∠ABC + ∠BOC

⇒ ∠ABD = ∠BOC

Thus, the external angle adjacent to ∠ABC is equal to ∠BOC.

Answer:

It is given that

We have to prove that.

Now in triangles and we have

(Given)

(Given)

So (common)

Each side of is equal to .

Hence, by congruence criterion we have Proved.

Answer:

It is given that 

and L, M, N are the mid points of sides, , and respectively.

We have to prove that

Now using the mid point theorem, we have

And

Similarly we have 

In triangle and we have

(Proved above)

(Proved above)

And (common)

So, by congruence criterion, we have 

And

Then

HenceProved.

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

Answer:

It is given that 

, and

If and

We have to prove that

In triangles and we have

(Since given)

So

And (given)

So by right hand side congruence criterion we have

So

HenceProved.

Answer:

In quadrilateral ABCD, AB = BC and AD = CD.

In ∆ABD and ∆CBD,

AB = CB        (Given)

BD = BD        (Common)

AD = CD        (Given)

∴ ∆ABD ≅ ∆CBD       (SSS congruence criterion)

So, ∠ABD = ∠CBD       .....(1)       (CPCT)

∠ADB = ∠CDB             .....(2)       (CPCT)

From (1) and (2), we conclude that

BD bisects both ∠ABC and ∠ADC.

Answer:

It is given ABCD is a square and

We have to prove that and

In right angled trianglesand Δ we have

And, and 

So by right hand side congruence criterion we have 

So (since triangle is congruent)

HenceProved.

Answer:

In quadrilateral ABCD, AB = AD and BC = CD. Let AC and BD intersect at O.

In ∆ABC and ∆ADC,

AB = AD        (Given)

AC = AC        (Common)

BC = CD        (Given)

∴ ∆ABC ≅ ∆ADC       (SSS congruence criterion)

⇒ ∠BAC = ∠DAC       (CPCT)

Now, in ∆ABO and ∆ADO,

AB = AD                   (Given)

∠BAO = ∠DAO        (Proved above)

AO = AO                   (Common)

∴ ∆ABO ≅ ∆ADO       (SAS congruence axiom)

⇒ OB = OD             .....(1)      (CPCT)

∠AOB = ∠AOD       .....(2)      (CPCT)

Now,

∠AOB + ∠AOD = 180º         (Linear pair)

⇒ 2∠AOB = 180º          [Using (2)]

⇒ ∠AOB = 90º        .....(3)

From (1) and (3), we conclude that AC is the perpendicular bisector of BD.

Answer:

O is a point in the interior of square ABCD. ∆OAB is an equilateral triangle.

Now,

∠DAB = ∠CBA        .....(1)         (Measure of each angle of a square is 90º)

∠OAB = ∠OBA        .....(2)         (Measure of each angle of an equilateral triangle is 60º)

Subtracting (2) from (1), we get

∠DAB − ∠OAB = ∠CBA − ∠OBA

⇒ ∠OAD = ∠OBC  

In ∆OAD and ∆OBC,

OA = OB      (Sides of an equilateral triangle are equal)

∠OAD = ∠OBC          (Proved above)

AD = BC       (Sides of a square are equal)

∴ ∆OAD ≅ ∆OBC       (SAS congruence axiom)

⇒ OD = OC             (CPCT)

In ∆OCD,

OC = OD

∴ ∆OCD is an isosceles triangle.    (A triangle whose two sides are equal is an isosceles triangle)

Answer:

ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.

Join AN and BN.

In ∆AMN and ∆BMN,

AM = BM                      (M is the mid-point of AB)

∠AMN = ∠BMN           (MN ⊥ AB)

MN = MN                      (Common)

∴ ∆AMN ≅ ∆BMN        (SAS congruence axiom)

So, AN = BN         .....(1)       (CPCT)

∠ANM = ∠BNM            (CPCT)

Now,

∠DNM = ∠CNM            (90º each)

∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM

⇒ ∠AND = ∠BNC      .....(2)

In ∆AND and ∆BNC,

DN = CN                       (N is the mid-point of CD)

∠AND = ∠BNC           [From (2)]

AN = BN                      [From (1)]

∴ ∆AND ≅ ∆BNC        (SAS congruence axiom)

So, AD = BC                  (CPCT)

Hence proved.

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

Answer:

In the triangle ABC it is given that 

We have to find the longest and shortest side.

Here

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence will be the longest

Since is the shortest angle so that side in front of it will be the shortest.

And is shortest side

Hence Is longest and is shortest.

Answer:

In the triangle ABC it is given that 

We have to find the longest side.

Here

(Since)

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence BC will be the longest side.

Answer:

It is given that

, and

We have to prove that 

(1)

(2)

(1) 

Now

And since BD=BC, so, and 

That is,

Now

And, so

Hence (1) (Side in front of greater angle will be longer)

And (2) Proved.

Answer:

As we know that a triangle can only be formed if

The sum of two sides is greater than the third side.

Here we have 2 cm, 3 cm and 7 cm as sides.

If we add

(Since 5 is less than 7)

Hence the sum of two sides is less than the third sides 

So, the triangle will not exist.

Answer:

It is given that

AP is the bisector of

We have to arrange, and in descending order.

In we have

(As AP is the bisector of)

So (Sides in front or greater angle will be greater)              ........(1)

In we have

(As AP is the bisector of)

Since, 

So                   ..........(2)

Hence 

From (1) & (2) we have

Answer:

(i) We have, quadrilateral ABCD

In $△\mathrm{ABC}$
$\mathrm{AB}+\mathrm{BC}>\mathrm{AC} \left(\mathrm{Triangle} \mathrm{inequality} \mathrm{property}\right) .....\left(1\right)$
Also, In $△\mathrm{DAC}$
$\mathrm{DA}+\mathrm{DC}>\mathrm{AC} \left(\mathrm{Triangle} \mathrm{inequality} \mathrm{property}\right) .....\left(2\right)$
Adding (1) and (2), we get
$\mathrm{DA}+\mathrm{DC}+\mathrm{AB}+\mathrm{BC}>2\mathrm{AC}$
Hence proved.

(ii) 
In $△\mathrm{ADC}$
$\mathrm{AD}+\mathrm{DC}>\mathrm{AC} \left(\mathrm{Triangle} \mathrm{inequality} \mathrm{property}\right) .....\left(1\right)$
Also, In $△\mathrm{ABC}$
$\mathrm{AB}+\mathrm{AC}>\mathrm{BC} \left(\mathrm{Triangle} \mathrm{inequality} \mathrm{property}\right) .....\left(2\right)$
Adding (1) and (2), we get
$\mathrm{AD}+\mathrm{DC}+\mathrm{AB}+\mathrm{AC}>\mathrm{AC} +\mathrm{BC}$
Subtracting AC from both sides, we get
$\mathrm{AD}+\mathrm{DC}+\mathrm{AB}>\mathrm{BC}$
Hence proved.

Answer:

It is given that, D is any point on side AC of a ∆ABC with AB = AC.



In ∆ABC,

AB = AC      (Given)

∴ ∠ACB = ∠ABC         (In a triangle, equal sides have equal angles opposite to them)

Now, ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC            (∠ACB = ∠ABC)

In ∆BCD,

∠DCB > ∠DBC

⇒ BD > CD                       (In a triangle, greater angle has greater side opposite to it)

Or CD < BD

Answer:

It is given that, is any point in the interior of

We have to prove that

(1) Produced to meet at.

In we have

    .........(1)

And in we have

            .........(2)

Adding (1) & (2) we get

HenceProved.

(2) We have to prove that

From the first result we have 

       ..........(3)

And

     .........(4)

Adding above (4) equation

HenceProved.

(3) We have to prove that

In triangles, and we have

Adding these three results

HenceProved.

Answer:

Let us consider $△\mathrm{ABC}$ where AC is the longest side.

Since AC is the longest side
$$\begin{gathered} \Rightarrow \mathrm{AC}>\mathrm{AB} \\ \Rightarrow \angle \mathrm{B}>\angle \mathrm{C} .....\left(1\right) \end{gathered}$$
We also know
$$\begin{gathered} \Rightarrow \mathrm{AC}>\mathrm{BC} \\ \Rightarrow \angle \mathrm{B}>\angle \mathrm{A} .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} \Rightarrow \angle \mathrm{B}+\angle \mathrm{B}>\angle \mathrm{A}+\angle \mathrm{C} \\ \Rightarrow 2\angle \mathrm{B}>\angle \mathrm{A}+\angle \mathrm{C} \end{gathered}$$
Adding $\angle B$ on both sides, we get
$$\begin{gathered} \Rightarrow 2\angle \mathrm{B}+\angle \mathrm{B}>\angle \mathrm{A}+\angle \mathrm{C}+\angle \mathrm{B} \\ \Rightarrow 3\angle \mathrm{B}>180° \\ \Rightarrow \angle \mathrm{B}>60° \\ \Rightarrow \angle \mathrm{B}>\frac{2}{3}\times \left(90°\right) \\ \Rightarrow \angle \mathrm{B}>\frac{2}{3} \mathrm{of} \mathrm{right} \mathrm{angle} \end{gathered}$$
Thus, it is proved that in a triangle, other than an equilateral triangle, the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer:

It is given that

Since, the triangles ABC and DEF are congruent, therefore,

Answer:

It is given that

For necessarily triangle to be congruent, sides should also be equal.

Answer:

It is given that

Since, two sides and angle between them are equal, therefore triangle ABC and DEF are congruent.

Answer:

The given information and corresponding figure is given below

From the figure, we have

And,

Hence, triangles ABC and ADC are congruent to each other.

Answer:

For the triangles ABC and ECD, we have the following information and corresponding figure:

In triangles ABC and ECD, we have

The SSA criteria for two triangles to be congruent are being followed. So both the triangles are congruent.

Answer:

In the triangle ABC it is given that 

, and are medians.

We have to show that

To show we will show that

In triangle ΔBFC and ΔBEC

As, so 

            .........(1)

BC=BC (common sides)   ........(2)

Since,

As F and E are mid points of sides AB and AC respectively, so

BF = CE         ..........(3)

From equation (1), (2), and (3)

HenceProved.

Answer:

In equilateral triangle we know that each angle is equal

So

Now (by triangle property)

Hence.

Answer:

We have to prove that

Given is a square

So

Now in is equilateral triangle.

So

In and

(Side of triangle)

(Side of equilateral triangle)

And,

So

Hence from congruence Proved.

Answer:

We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.

In we have

, and

We have to prove 

As we know perpendicular line segment is shortest in length

Since

So     ........(1)

And 

       ........(2)

Adding (1) and (2) we get

   ........(3)

Now, so

       .......(4)

And again, this implies that

    ........(5)

Adding (3) & (4) and (5) we have

HenceProved.

Answer:

It is given that

, and

We have to prove that

We basically will prove to show

In and

(Given)

(Given)

And is common in both the triangles

So all the properties of congruence are satisfied

So

Hence Proved.

Answer:

It is given that, AB = QR, BC = PR and CA = PQ.

So, A ↔ Q, B ↔ R and C ↔ P

∴ ∆ABC ≅ ∆QRP

If AB = QR, BC = PR and CA = PQ, then triangle __ABC__ ≅ triangle __QRP__.

Answer:

SAS congruence axiom states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.

In ∆ABC, ∠A is included between the sides AB and AC.

In ∆DEF, ∠D is included between the sides DF and DE.



∴ ∆ABC ≅ ∆DEF by SAS axiom if AC = DE.

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __AC = DE__.

Answer:

If two sides of a triangle are equal, then it is an isosceles triangle.

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

In ∆ABC,

AB = AC    (Given)

∴ ∠C = ∠B       (In a triangle, equal sides have equal angles opposite to them)

It is given that, ∠C = ∠P and ∠B = ∠Q.

∴ ∠P = ∠Q

In ∆PQR,

∠P = ∠Q        (Proved)

∴ QR = PR     (In a triangle, equal sides have equal angles opposite to them)

So, ∆PQR is an isosceles triangle.

However, it cannot be proved that the corresponding sides of ∆ABC are congruent to the corresponding sides of ∆PQR. Hence, the triangles are not congruent.

In ∆ABC and ∆PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __isosceles__ but not __congruent__.

Answer:

In ∆PQR,

∠P = ∠R       (Given)

∴ QR = PQ    (Sides opposite to the equal angles of a triangle are equal)

⇒ PQ = 4 cm       (QR = 4 cm)

In ∆PQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ___4 cm___.

Answer:

In ∆ABC,

AB = AC           (Given)

∴ ∠C = ∠B        (Angles opposite to the equal sides of a triangle are equal)

⇒ ∠C = 50°       (∠B = 50°)

In ∆ABC, AB = AC and ∠B = 50°. Then, ∠C = __50°__.

Answer:

In ∆ABC,

AB = BC           (Given)

∴ ∠C = ∠A        ..... (1)         (Angles opposite to the equal sides of a triangle are equal)

Now,

∠A + ∠B + ∠C = 180°       (Angle sum property of triangle)

∴ ∠A + 80° + ∠A = 180°

⇒ 2∠A = 180° − 80° = 100°

⇒ ∠A = $\frac{100°}{2}$ = 50°

In ∆ABC, AB = BC and ∠B = 80°. Then ∠A = __50°__.

Answer:

In ∆PQR,

$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180°$          (Angle sum property of a triangle)

$\therefore 70°+\angle \mathrm{Q}+30°=180°$

$\Rightarrow \angle \mathrm{Q}+100°=180°$

$\Rightarrow \angle \mathrm{Q}=180°-100°=80°$



So, ∠Q is the greatest angle in the ∆PQR.

We know that, in a triangle the greater angle has the longer side opposite to it.

∴ PR is the longest side of ∆PQR.

If in ∆PQR, ∠P = 70° and ∠R = 30°, then the longest side of ∆PQR is ___PR___.

Answer:

In ∆PQR,

∠R > ∠Q        (Given)

∴ PQ > PR     (In a triangle, the greater angle has the longer side opposite to it)

In ∆PQR, if ∠R > ∠Q, then ___PQ > PR___.

Answer:

In ∆ABC, AD is the bisector of ∠A.

∴ ∠BAD = ∠CAD             .....(1)

We know that exterior angle of a triangle is greater than each of interior opposite angle.

In ∆ABD,

∠ADC >  ∠BAD

⇒ ∠ADC > ∠CAD        [Using (1)]

In ∆ADC,

∠ADC > ∠CAD

∴ AC > CD           (In a triangle, the greater angle has the longer side opposite to it)

Similarly, AB > BD

D is a point on side BC of a ∆ABC such that AD bisects ∠BAC. Then ___AC > CD and AB > BD___.

Answer:

In a triangle, the sum of two sides is greater than the third side.

∴ 5 cm + 1.5 cm > Third side

Or Third side < 6.5 cm

In a triangle, the difference of two sides is less than the third side.

∴ 5 cm − 1.5 cm < Third side

Or Third side > 3.5 cm

So,

3.5 cm < Third side < 6.5 cm

Thus, the length of the third side of the triangle lies between 3.5 cm and 6.5 cm.

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between __3.5 cm__ and _6.5 cm _.
 

Answer:

AD is the median of the ∆ABC.

In ∆ABD,

AB + BD > AD      .....(1)          (In a triangle, the sum of any two sides is greater than the third side)

In ∆ACD,

CD + CA > AD      .....(2)          (In a triangle, the sum of any two sides is greater than the third side)

Adding (1) and (2), we get

AB + BD + CD + CA > AD + AD

⇒ AB + BC + CA > 2AD                     (BC = BD + CD)

⇒ Perimeter of ∆ABC > 2AD

Or the perimeter of ∆ABC cannot be less than or equal to 2AD

If AD is a median of ∆ABC, then the perimeter of ∆ABC cannot be less than or equal to ___2AD___.

Answer:

If ∆PQR ≅ ∆EDF, then

P ↔ E, Q ↔ D and R ↔ F 

So, PR = EF,  PQ = ED and QR = DF       (If two triangles are congruent, then their corresponding sides are congruent)

If ∆PQR ≅ ∆EDF, then PR = ____EF____.

Answer:

If ∆ABC ≅ ∆RPQ, then

A ↔ R, B ↔ P and C ↔ Q

So, AB = RP, BC = PQ and CA = QR       (If two triangles are congruent, then their corresponding sides are congruent)

It is given that ∆ABC ≅ ∆RPQ, then BC = ___PQ___.

Answer:

In ∆ABC and ∆PQR,

∠A = ∠Q

∠B = ∠R

AC = PR

∴ ∆ABC ≅ ∆QRP      (AAS congruence criterion)

AAS congruence criterion states that if any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

In ∆ABC and ∆PQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by __AAS congruence__ criterion.

Answer:

In ∆ABC and ∆PQR,

∠A = ∠Q

∠B = ∠R

AB = QR

∴ ∆ABC ≅ ∆QRP      (ASA congruence criterion)

ASA congruence criterion states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

In ∆ABC and ∆PQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by __ASA congruence criterion__.