## MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.

∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. below). If AD is extended to intersect BC at P, show that

- ∆ABD = ∆ACD
- ∆ABP = ∆ACP
- AP bisects ∠A as well as ∠D.
- AP is the perpendicular bisector of BC.

Solution:

Given

AB = AC

DB = DC

To prove:

- ∆ABD = ∆ACD
- ∆ABP = ∆ACP
- ∠1 = ∠2 and ∠5 = ∠6
- ∠3 = ∠4 = 90° and BP = PC

Proof

1. In ∆ABD and ∆ACD

AB = AC (given)

BD = CD (given)

AD =AD (common)

∆ABD = ∆ACD (by SSS)

and so ∠1 = ∠2 (by CPCT)

2. In ∆ABP and ∆ACP

AB = AC (given)

∠1 = ∠2 (proved)

AP = AP (common)

∆ABP ≅ ∆ACP (by SAS)

and so ∠3 = ∠4

and BP = CP (by CPCT)

∠1 = ∠2 AP bisects ∠A

3. In ∆DBP and ∆DCP

BP = CP (proved)

∠3 = ∠4 (proved)

DP = DP (common)

∆DBP ≅ ∆DCP (by SAS)

and so ∠5 = ∠6 (by CPCT)

∴ AP bisects ∠D.

4. BP = CP (proved)

∠3 = ∠4 (proved)

∠3 + ∠4 = 180° (LPA’s)

∠3 + ∠3 = 180° (∠3 = ∠4)

2∠3 = 180°

∠3 = 90°

∠3 = ∠4 (each 90°)

and therefore, AP is the perpendicular bisector of BC

Question 2.

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

- AD bisects BC
- AD bisects ∠A.

Solution:

Given

AB = AC and AD ⊥ BC

To prove

∠1 = ∠2 and

BD = CD

Proof:

In ∆ABD and ∆ACD

AB = AC

AD = AD

∠3 = ∠4

∆ABD = ∆ADC

and so BD = CD and ∠1 = ∠2

Question 3.

Two sides AS and BC and median AM on one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see-Fig. below). Show that:

- ∆ABM ≅ ∆PQN
- ∆ABC ≅ ∆PQR

Solution:

Given

AB = PQ

BC = QR

AM = PN

To prove:

- ∆ABM ≅ ∆PQN
- ∆ABC ≅ ∆PQR

Proof:

BC = QR (given)

\(\frac{1}{2}\) BC = \(\frac{1}{2}\)QR

BM = QN

1. In ∆ABM and ∆PQN

AB = PQ (given)

BM = QN (proved)

AM = PN (given)

∆ABM ≅ ∆PQN (by SSS)

and so ∠ABC = ∠PQR (by CPCT)

2. In ∆ABC and ∆PQR

AB = PQ (given)

BC = QR (given)

∠B = ∠Q (proved)

∆ABC ≅ ∆PQR by (SAS)

Question 4.

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that triangle ABC is isosceles.

Solution:

Given

∠E = ∠F

BE = CF

To prove

AB = AC

Proof:

In ∆FBC and ∆ECB

BE = CF (given)

∠F = ∠E (each 90°)

BC = CB (common)

∴ ∆FBC = ∆ECB (by RHS)

and so ∠B = ∠C (by CPCT)

In ∆ABC, ∠B = ∠C

AB = AC

(sides opposite to equal angles of a A are equal)

and so ABC is isosceles.

Question 5.

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:

Given

AB = AC

∠APB = ∠APC = 90°

To prove:

∠B = ∠C

Proof:

In ∆ABP and ∆ACP

AP = AP (common)

∠APB = ∠APC (each 90°)

AB = AC (given)

∴ ∆ABP = ∆ACP (by RHS)

and ∠B = ∠C (by CPCT)

Theorem 7.6.

If two angles of a triangle are equal, then the sides opposite to them are also equal.

Given

In ∆ABC, ∠C = ∠B

To prove: AB = AC

Construction:

Draw the bisector of ∠A and let it meet BC an D

Proof:

In ∆s ABD and ACD, we have ∠B = ∠C (Given)

∠BAD = ∠CAD (Construction)

AD = AD (Common)

∴ ∆ABD ≅ ∆ACD (AAS Cong. Criterion)

Hence, AB = AC (CPCT)

Theorem 7.7.

If two sides of a triangle are unequal, the longest side has greater angle opposite to it.

Given:

In ∆ABC; AC > AB

To prove: ∠ABC > ∠ACB.

Construction:

Mark point D on AC such that AB = AD. Join BD.

Proof:

In ∆ABD,

AB = AD

∴ ∠1 = ∠2 (Const. As opp. equal sides) ….(1)

But ∠2 is an exterior angle of ABCD.

∠2 > ∠ACB (Exterior Angle Theorem) …(2)

From (1) and (2), we have

∠1 > ∠ACB (Const.)

But ∠ABC > ∠1

∴ ∠ABC > ∠ACB

Theorem 7.8. (Converse of Theorem 7.7)

In a triangle the greater angle has the longer side opposite to it.

Given:

In ∆ABC, ∠ABC > ∠ACB

To prove: AC > AB

Proof:

For ∆ABC, there are only three possibilities of which exactly one must be true.

- AC = AB
- AC < AB (iii) AC > AB.

Case 1:

If AC = AB, then ∠ABC = ∠ACB, which is contrary to what is given.

AB ≠ AC

Case 2:

If AC < AB, the longer side has the greater angle opposite to it.

∴ ∠ACB > ∠ABC.

This is also contrary to what is given.

Case 3:

We are left with the only possibility, namely AC > AB which is true.

AC > AB.