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Question
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ
Solution
In ∆PQT and ∆STR we have
∠P = ∠S = 90° ...(Given)
∠PTQ = ∠STR ...(Vertically opposite angle)
By AA similarity
∆PTQ ~ ∆STR we get
`"PT"/"ST" = "TQ"/"TR"`
PT × TR = ST × TQ
Hence it is proved.
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