Question

In triangle PQR, A and B are two points on QR such that
angle PAB = angle PBA and QA = BR. Prove that triangle PQA is congruent to triangle PRB.

please explain it step by step

see the diagram from the photo​

Answer 1

Answer:

Hope it will be helpful to you

Step-by-step explanation:

Here, in △PQR, A and B are points on side QR such that QA = AB = BR.

Through P, draw a line l parallel to QR

Now, △PQA, △PAB and △PBR on the equal bases and between same parallels l || QR

⇒ ar(△PQA) = ar(△PAB) = ar(△PBR) ---i)

Now, ar(PQB) = ar(△PQA) + ar(△PAB)

= 2ar(△PQA) [using (i)]

= 2ar(△PBR) [using (ii)]

Answer 2

$Answer⬇️$

In ∆ABC and ∆PQR,

AB/PQ = BC/QR

⇒ AB/BC = PQ/QR

Angle formed by sides AB and BC is ∠B

Angle formed by sides PQ and QR is ∠Q

∴ ∠B = ∠Q is essential for similarity of two triangles. thus ∠B = ∠Q

AB/BC = PQ/QR

∴ ∆ABC ~ ∆PQR (SAS similarity theorem)