RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

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.