Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 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 figure). If AD is extended to intersect BC at P, show that:

(i) ∆ABD ≅ AACD

(ii) ∆ABP ≅ AACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Answer:

(i) In ∆s ABD and ACD, we have :

AB = AC (Given)

BD = CD (Given)

and AD = AD (Common)

∴ By SSS criterion of congruence, we have:

∆ABD ≅ ∆ACD.

(ii) In ∆s ABP and ACP, we have :

AB = AC (Given)

∠BAP = ∠CAP

[∵ ∆ABD ≅ ∆ACD ⇒ ∠BAD = ∠DAC ⇒ ∠BAP = ∠CAP (CPCT)]

and AP = AP (Common)

∴ By SAS criterion of congruence, we have :

∆ABP ≅ ∆ACP.

(iii) Since ∆ABD ≅ ∆ACD, therefore

∠BAD = ∠CAD …….. (1)

⇒ AD bisects ZA

⇒ AP bisects ZA.

Now, in ∆s BDP and CDP, we have:

BD = CD (Given)

BP = CP (∵ ∆ABP ≅ ∆ACP ⇒ BP = CP)

and DP = DP (Common)

∴ By SSS criterion of congruence, we have:

∆BDP ≅ ∆CDP

∴ ∠BDP = ∠CDP

⇒ DP bisects ∠D ⇒ AP bisects ∠D. ……… (2)

Combining (1) and (2), we get:

AP bisects ∠A as well as ∠D.

(iv) Since AP stands on BC

∴ ∠APB + ∠APC = 180° (Linear pair)

But ∠APB = ∠APC (∵ ∆APB ≅ ∆APC ⇒ ∠APB = ∠APC)

∴ ∠APB = ∠APC = \(\frac{180^{\circ}}{2}\) = 90°.

Also, BP = CP (Proved above)

So, AP is perpendicular bisector of BC.

Hence proved.

Question 2.

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

(i) AD bisects BC

(ii) AD bisects ZA.

Answer:

AD is the altitude drawn from vertex A of an isosceles ∆ABC to the opposite base BC so that AB = AC,

∠ADC = ∠ADB = 90°.

Now, in As ADB and ADC, we have :

Hyp. AB = Hyp. AC (Given)

AD = AD (Common)

and ∠ADB = ∠ADC

∴ By RHS criterion of congruence, we have:

∆ADB ≅ ∆ADC

∴ BD = DC and ∠BAC = ∠DAC

(∵ Corresponding parts of congruent triangles are equal.)

Hence, AD bisects BC, which proves (i), and AD bisects ∠A, which proves (ii).

Question 3.

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see figure). Show that:

(i) ∆ABM ≅ ∆PQN

(ii) ∆ABC ≅ ∆PQR

Answer:

Two ∆s ABC and PQR in which AB = PQ, BC = QR and AM = PN.

Since AM and PN are medians of As ABC and PQR respectively.

Now, BC = QR

BM = QN

Now, in ∆s ABM and PQN, we have :

AB = PQ (Given)

BM = QN [From (1)]

and, AM = PN (Given)

∴ By SSS criterion of congruence, we have:

∆ABM ≅ ∆PQN, which proves (i)

So, ∠B = ∠Q ………. (2)

(∵ Corresponding parts of congruent triangles (CPCT) are equal.)

Now, in ∆s ABC and PQR, we have :

AB = PQ (Given)

∠B = ∠Q [From (2)]

and BC = QR (Given)

∴ By SAS criterion of congruence, we have:

∆ABC ≅ ∆PQR, which proves (ii).

Hence proved.

Question 4.

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

Answer:

In ∆s BCF and CBE, we have :

∠BFC = ∠CEB

Hyp. BC = Hyp. BC

CF = BE

∴ By RHS criterion of congruence, we have :

∆BCF ≅ ∆ CBE

So, ∠FBC = ∠ECB

(∵ Corresponding parts of congruent triangles are equal.)

Now, in ∆ABC, ∠ABC = ∠ACB (∵ ∠FBC = ∠ECB)

∴ AB = AC

(∵ Sides opposite to equal angles of a triangle are equal.)

∴ ∆ABC is an isosceles triangle.

Hence proved.

Question 5.

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

Answer:

In ∆s ABP and ACP, we have:

AB = AC (Given)

AP = AP (Common)

and ∠APB = ∠APC (∵ Each = 90°)

∴ By RHS criterion of congruence, we have:

∆ ABP ≅ ∆ ACP

So, ∠B = ∠C

(∵ Corresponding parts of congruent triangles are equal.)

Hence Proved.

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