Question

In the given figure, RQ and TP are perpendiculars to PQ and TS perpendicular to PR. Prove that
ST.RQ = PS.PQ.
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Answer 1

In ∆RQP

By angle sum property

→ ∠R + ∠P + ∠Q = 180°

→ ∠1 + ∠2 + ∠3 = 180°

Given that, ∠3 = 90°. So,

→ ∠1 + ∠2 + 90° = 180°

→ ∠1 + ∠2 = 90°.....(eq 1)

TS is a perpendicular drawn on line PR such that ∠TSP = ∠5 = 90°

Also, TP is perpendicular on line PQ such that ∠TPQ = 90°

→ ∠4 + ∠2 = 90°.....(eq 2)

On comparing (eq 1) and (eq 2) we get,

→ ∠1 + ∠2 = ∠4 + ∠2

→ ∠1 = ∠4

In ∆RQP and ∆TSP

→ ∠3 = ∠5 (each 90°)

→ ∠1 = ∠4 (from above)

By AA

∆RQP ~ ∆TSP

Therefore,

→ ST/PQ = PS/RQ

Cross -multiply them

→ ST.RQ = PS.PQ

Hence, proved

Answer 2

||✪✪ QUESTION ✪✪||

In the given figure, RQ and TP are perpendiculars to PQ and TS perpendicular to PR. Prove that ST.RQ = PS.PQ.

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

Concept used :-

→ AA Similarity Theorem :- If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

→ When two ∆'s are similar, the ratios of the lengths of their corresponding sides are equal.

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