Question

In the given figure, PS and QT are perpendiculars to QR and PR respectively. Then, prove that triangle RST is similar to Triangle RPQ

Answer 1

Solution :-

Draw a circle taking quadrilateral PQST inside it .

As we can see that,

→ ∠PTQ = ∠PSQ { given each 90° }

since both angles are drawn on circumference by chord PQ and they are equal in measure . Therefore , we can conclude that, quadrilateral PQST is a cyclic quadrilateral .

now,

→ ∠PTS + ∠SQP = 180° { sum of opposite angles of a cyclic quad. are equal to 180° . }

→ ∠SQP = 180° - ∠PTS ----------- Eqn.(1)

now,

→ ∠RTS + ∠PTS = 180° { RP is a straight line }

→ ∠RTS = 180° - ∠PTS ----------- Eqn.(2)

from Eqn.(1) and Eqn.(2) we get,

→ ∠SQP = ∠RTS

→ ∠RQP = ∠RTS --------- Eqn.(3)

now, in ∆RST and ∆RPQ we have,

→ ∠RTS = ∠RQP { from Eqn.(3) }

→ ∠TRS = ∠QRP { common }

hence,

→ ∆RST ~ ∆RPQ { By AA similarity. } (Proved)

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