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

1.	In figure, if ∠1 =∠2 and ΔNSQ = ΔMTR, then prove that ΔPTS ~ ΔPRQ.
Answer» According to the question, ∆ 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 ΔNSQ = ΔMTR, then prove that ΔPTS ~ ΔPRQ.

Answer»

According to the question,

∆ 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.

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