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Question 13
In the given figure, l||m and line segments AB, CD and EF are concurrent at point P. Prove that
AEBF=ACBD=CEFD.

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Solution

Given IIIm and line segments AB, CD and EF are concurrent at point P.

To prove that AEBF=ACBD=CEFD
Proof
In ΔAPC and ΔBPD,
APC=BPD [vertically opposite angles]
PAC=PBD [alternate angles]
ΔAPCΔBPD [by AAA similarity criterion]
Then, APPB=ACBD=PCPD ...(i)

In ΔAPE and ΔBPF,
APE=BPF [vertically opposite angles]
PAE=PBF [ alternate angles]
ΔAPEΔBPF [by AAA similarity criterion]
Then, APPB=AEBF=PEPF …..(ii)

In ΔPEC and ΔPFD,
EPC=FPD [vertically opposite angles]
PCE=PDF [ alternate angles]
ΔPECΔPFD [by AAA similarity criterion]
Then, PEPF=PCPD=ECFD ….(iii)
From, Eq.(i), (ii) and (iii),
APPB=ACBD=AEBF=PEPF=ECFD
AEBF=ACBD=CEFD


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