RD Sharma 2020 solution class 9 chapter 12 Congruent Triangles Exercise 12.3

Exercise 12.3

Page-12.47

Question 1:

In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.

Answer 1:

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.

Question 2:

If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.

Answer 2:

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.

Question 3:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Answer 3:

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 

Question 4:

Prove that each angle of an equilateral triangle is 60°

Answer 4:

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.

Question 5:

Angles A, B, C of a triangle ABC are equal to each other. Prove that ΔABC is equilateral.

Answer 5:

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.

Question 6:

ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer 6:

It is given that 

We have to find and.

Since so,

Now (property of triangle)

(Since )

Here

Then

Hence 

Question 7:

PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.

Answer 7:

It is given that

We have to prove

In we have 

(Given)

So,

Now (Given)

Since corresponding angle are equal, so

That is, 

Henceproved.

Question 8:

In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced  prove that ∠MOC = ∠ABC.

Answer 8:

It is given that

In,

We have to prove that

Now 

(Given)

Thus

   ........(1)

In, we have

So, {from equation (1)}

Hence Proved.

Question 9:

P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

Answer 9:

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.

Question 10:

ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.

Answer 10:

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.

Question 11:

Bisectors of angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.

Answer 11:

∠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.