Question

In ∆PQR,PS is perpendicular to QR, QT is perpendicular to PR. prove that ∆QTR~∆PSR​

Answer 1

Answer:

your question is wrung

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

In right angled ∆PSQ & ∆PSR

PQ²= PS²+QS²…………(1)

and PR²= PS²+SR²………(2)

On adding eq 1 & 2

PQ²+PR²= 2PS²+QS²+SR²

PQ²+PR²= 2(QS×SR) + QS²+SR²

[ GIVEN: PS²= QS×RS]

PQ²+PR²=(QS+SR)²

[a²+b²+2ab= (a+b)²]

PQ²+PR²=QR²        [ QR= QS+SR]

Hence,∆PQR is a right angled ∆ right angled at P.

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