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.
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]
∴ ΔAPC∼ΔBPD [by AAA similarity criterion]
Then, APPB=ACBD=PCPD ...(i)
In ΔAPE and ΔBPF,
∠APE=∠BPF [vertically opposite angles]
∠PAE=∠PBF [ alternate angles]
ΔAPE∼ΔBPF [by AAA similarity criterion]
Then, APPB=AEBF=PEPF …..(ii)
In ΔPEC and ΔPFD,
∠EPC=∠FPD [vertically opposite angles]
∠PCE=∠PDF [ alternate angles]
∴ ΔPEC∼ΔPFD [by AAA similarity criterion]
Then, PEPF=PCPD=ECFD ….(iii)
From, Eq.(i), (ii) and (iii),
APPB=ACBD=AEBF=PEPF=ECFD
∴ AEBF=ACBD=CEFD