In figure, if ∠ 1 = ∠ 2 and ∆NSQ ≅ ∆MTR, then prove that ∆PTS ∼ ∆PRQ.

Question

Philoid · Accepted Answer

Given: ∆ NSQ ≅ ∆MTR

∠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 Equation (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.

In figure, if ∠1 = ∠2 and ∆NSQ ≅ ∆MTR, then prove that ∆PTS ∼ ∆PRQ.