In figure, if ∠1 =∠2 and ΔNSQ ≅ ΔMTR, then prove that ΔPTS ~ ΔPRQ.
ΔNSQ ≅ ΔMTR
∠1 = ∠2
∆NSQ = ∆MTR
SQ = TR ......(i)
∠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
∠2 = ∠PRQ
In ∆PTS and ∆PRQ.
∠P = ∠P ......[Common angles]
∠1 = ∠PQR .....(Proved)
∠2 = ∠PRQ ....(Proved)
∴ ∆PTS – ∆PRQ ......[By AAA similarity criteria]
Concept: Basic Proportionality Theorem (Thales Theorem)
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