Question

AC||BD IS AE/CE=DE/BE

Answer 1

given,

in trapezium ABCD, diagonals AD and BC intersect each other at E

to prove, AE/CE = DE/BE

construction : from E a line parallel to AC is drawn which intersect AB at F.

proof : in triangle BAC,    EFllAC

∴ BF/FA = BE/EC ( by bpt )  ----------- (1)

again,     in triangle ABD,  EFllBD

∴ BF/FA = DE/EA (by bpt) --------------(2)

from (1) and (2) we get,

BE/EC = DE/EA

⇒ AE/EC = DE/BE.....

                                     hence proved.