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Question
In the following figure, `("QR")/("QS") = ("QT")/("PR")` and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Solution
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR ...(i)
Given,
`("QR")/("QS") = ("QT")/("PR")`
Using (i), we get
`("QR")/("QS") = ("QT")/("QP")` ...(ii)
In ΔPQS and ΔTQR,
`("QR")/("QS") = ("QT")/("QP")`
∠Q = ∠Q = 1
∴ ΔPQS ~ ΔTQR ...[SAS similarity criterion]
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