Question

In figure, if ∠1 = ∠2 and ΔNSQ ≅ ΔMTR, then prove that ΔPTS ~ ΔPRQ.

Answer 1

According to the question,

ΔNSQ ≅ ΔMTR

∠1 = ∠2

Since,

∆NSQ = ∆MTR

So,

SQ = TR  ...(i)

Also,

∠1 = ∠2 ⇒ PT = PS  ...(ii) [Since, sides opposite to equal angles are also equal]

From equations (i) and (ii),

`("PS")/("SQ") = ("PT")/("TR")`

⇒ ST || QR

By converse of basic proportionality theorem, If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.

∴ ∠1 = PQR

And

∠2 = ∠PRQ

In ∆PTS and ∆PRQ,

∠P = ∠P   ...[Common angles]

∠1 = ∠PQR  ...(Proved)

∠2 = ∠PRQ   ...(Proved)

∴ ∆PTS – ∆PRQ   ...[By AAA similarity criteria]

Hence proved.