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Answer:

To, Solve this Question we first need draw a circle outside the triangle.

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 ( Equation 1 )

Now,

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

→ ∠RTS = 180° - ∠PTS (Equation 2)

from Equation (1) and Equation (2) we get,

→ ∠SQP = ∠RTS

→ ∠RQP = ∠RTS ( Equation 3 )

Now, in ∆RST and ∆RPQ we have,

→ ∠RTS = ∠RQP { from ( Equation 3) }

→ ∠TRS = ∠QRP { common }

hence,

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

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