Given IIIm and line segments AB, CD and EF are concurrent at point P.
To prove that AEBF=ACBD=CEFDProofIn 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