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Question
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
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Solution
Given: ∠QPR = 90°; PS is the bisector of ∠P. ST ⊥ ∠PR
To prove: ST × (PQ + PR) = PQ × PR
Proof: In ∆PQR, PS is the bisector of ∠P.
∴ `"PQ"/"QR" = "QS"/"SR"`
Adding (1) on both side
`1 + "PQ"/"QR" = 1 + "QS"/"SR"`
`("PR" + "PQ")/"PR" = ("SR"+ "QS")/"SR"`
`("PQ" + "PR")/"PR" = "QR"/"SR"` ...(1)
In ∆RST And ∆RQP
∠SRT = ∠QRP = ∠R ...(Common)
∴ ∠QRP = ∠STR = 90°
∆RST ~ RQP ...(By AA similarity)
`"SR"/"QR" = "ST"/"PQ"`
`"QR"/"SR" = "PQ"/"ST"` ...(2)
From (1) and (2) we get
`("PQ" + "PR")/"PR" = "PQ"/"ST"`
ST × (PQ + PR) = PQ × PR
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