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Question
In figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that `(AE)/(BF) = (AC)/(BD) = (CE)/(FD)`.
Solution
Given l || m and line segments AB, CD and EF are concurrent at point P.
To prove: `("AE")/("BF") = ("AC")/("BD") = ("CE")/("FD")`
Proof: In ΔAPC and ΔBPD,
∠APC = ∠BPD ...[Vertically opposite angles]
∠PAC = ∠PBD ...[Alternative angles]
∴ ΔAPC ∼ ΔBPD ...[By AA similarity criterion]
Then, `("AP")/("PB") = ("AC")/("BD") = ("PC")/("PD")` ...(i)
In ΔAPE and ΔBPF,
∠APE = ∠BPF ...[Vertically opposite angles]
∠PAE = ∠PBF ...[Alternative angles]
∴ ΔAPE ∼ ΔBPF ...[By AA similarity criterion]
Then, `("AP")/("PB") = ("AE")/("BF") = ("PE")/("PF")` ...(ii)
In ΔPEC and ΔPFD,
∠EPC = ∠FPD ...[Vertically opposite angles]
∠PCE = ∠PDF ...[Alternative angles]
∴ ΔPEC ∼ ΔPFD ...[By AA similarity criterion]
Then, `("PE")/("PF") = ("PC")/("PD") = ("EC")/("FD")` ...(iii)
From equations (i), (ii) and (iii),
`("AP")/("PB") = ("AC")/("BD") = ("AE")/("BF") = ("PE")/("PF") = ("EC")/("FD")`
∴ `("AE")/("BF") = ("AC")/("BD") = ("CE")/("FD")`
Hence proved.
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