## MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1

Question 1.

In quadrilateral ABCD, AC = AD and AB bisects ∠A (see below). Show that ∆ABC = ∆ABD. What can you say about BC and BD?

Solution:

Given AC = AD

AB is the bisector of ∠A

i. e., ∠1 = ∠2

To prove: ∆ABC = ∆ABD

Proof:

In ∆ABC and ∆ABD

∠1 = ∠2 (given)

AC = AD (given)

AB = AB (common)

∆ABC ≅ ∆ABD (by SAS)

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see below). Prove that

- ∆ABD = ∆BAC
- BD = AC
- ∠ABD = ∠BAC

Solution:

Given AD = BC

∠DAB = ∠CBA

To prove:

- ∆ABD ≅ ∆BAC
- BD = AC
- ∠ABD = ∠BAC

Proof:

1. In ∆ABD and ∆BAC

AD = BC (Given)

∠DAB = ∠CBA (given)

AB = BA (common)

∆ABD = ∆BAC (by SAS)

and so 2. BD = AC (by CPCT)

and so 3. ∠ABD – ∠BAC (by CPCT)

Question 3.

AD and BC are equal perpendiculars to a line segment AB (see below). Show that CD bisects AB.

Solution:

Given: ∠B = ∠A (each 90°)

AD = BC

To prove: OA = OB

Proof:

In ∆OBC and ∆OAD

∠B = ∠A (each 90°)

BC = AD (given)

∠BOC = ∠AOD (V.O.A’s)

∆OBC = ∆OAD (by SAS)

and so OA = OB (by CPCT)

Question 4.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see below). Show that ∆ABC = ∆CDM.

Solution:

Given: l ∥ m and p ∥ q

To prove:

∆ABC = ∆CDA

Proof:

In quadrilateral ABCD

AB ∥ DC and BC ∥ AD

∴ ABCD is a parallelogram

and so AB = DC [Opposite sides]

BC = AD

In ∆ABC and ∆CDA

AB = CD (proved)

BC = DA (proved)

AC = CA (proved)

∆ABC = ∆CDA (by SSS)

Question 5.

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see below). Show that

- ∆APB = ∆AQB
- BP = BQ or B is equidistant from the arms of ∠A.

Solution:

Given

∠1 = ∠2

BQ ⊥ AC

and BP ⊥ AD

To prove:

- ∆APB ≅ ∆AQB
- BP = BQ

Proof:

In ∆APB and ∆AQB

∠2 = ∠1 (given)

∠P = ∠Q (each 90°)

AB = AB (common)

∆APB ≅ ∆AQB (byAAS)

BP = BQ (by CPCT)

Question 6.

In Fig. given below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

Given

AC = AE

AB = AD

∠BAD = ∠EAC i.e., ∠1 = ∠2

To prove: BC = DE

Proof:

In ∆ABC and ∆ADE

AB = AD (given)

AC = AE (given)

∠1 = ∠2 (given)

Adding ∠3 on both sides

∠1 + ∠3 = ∠2 + ∠3

∠BAC = ∠DAE

∆ABC = ∆ADE , (by SAS)

and so BC = DE (by CPCT)

Question 7.

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see below). Show that

- ∆DAP ≅ ∆EBP
- AD = BE

Solution:

Given:

AP = BP

∠BAD = ∠ABE

∠EPA = ∠DPB

∠1 = ∠2

To prove:

- ∆DAP ≅ ∆EBP
- AD = BE

Proof:

∠1 = ∠2 (given)

Adding ∠3 on both sides

∠1 + ∠3 = ∠2 + ∠3

∠APD = ∠BPE

In ∆DAP and ∆EBP

∠APD = ∠BPE (proved)

AP = BP (given)

∠PAD = ∠PBE (given)

∆DAP ≅ ∆EBP (by ASA)

and so AD = BE (by CPCT)

Question 8.

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see below). Show that:

- ∆AMC = ∆BMD
- ∠DBC is a right angle.
- ∆DBC = ∆ACB
- CM = \(\frac{1}{2}\) AB

Solution:

∠C = 90°

AM = BM

DM = CM

To prove:

- ∆AMC ≅ ∆BMD
- ∆DBC = 90°
- ∆DBC ≅ ∆ACB
- CM = \(\frac{1}{2}\) AB

Proof:

1. ∆AMC and ∆BMD

AM = BM (given)

MC = MD (given)

∠1 = ∠2 (V.O.A’s)

∆AMC = ∆BMD (by SAS)

and so AC = DB and ∠4 and ∠3 (by CPCT)

2. ∠4 = ∠3 (proved) [AIA’s]

∴ DB ∥ AC

AC ∥ BD and BC is the transversal

∠C + ∠B = 180° (C.I.A’s)

∠B = 180° – 90° = 90°

3. In DBC and ∆ACB

DB = AC (proved)

∠B = ∠C (each 90°)

BC = CB (common)

∆DBC = ∆ACB (by SAS)

and so DC = AB (by CPCT)

4. DC =AB (proved)

\(\frac{1}{2}\)DC = \(\frac{1}{2}\)AB

CM = \(\frac{1}{2}\)AB.

Theorem 7.3

AAS (Angle-Angle-Side) Congruence Theorem:

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.

Given:

In ∆s ABC and DEF, we have

∠A = ∠D

∠B = ∠E

and BC = EF

To prove:

∆ABC = ∆DEF

Proof:

Since the sum of the angles of a triangle is 180°. We have

∠A + ∠B + ∠C = ∠D + ∠E + ∠F

Since ∠A = ∠D and ∠B – ∠E (Given)

∠C = ∠F …..(i)

Now, in ∆ABC and ∆DEF, we have

∠B = ∠E (Given)

∠C = ∠F [From, (i)]

and BC = EF (Given)

∴ ∆ABC ≅ ∆DEF [ASA Cong. Theorem]