Question 13 In the given figure, lIIm and line segments AB, CD and EF are concurrent at point P. Prove that AEBF=ACBD=CEFD.
Open in App
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] ∴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