Question

The side AB of the parallelogram ABCD is
produced to X and the bisector of angle CBX meets
DA produced and DC produced at E and F
respectively. Prove that DE = DF = AB + BC.​

Answer 1

CBF=FBX (given )

DC=AB ( ,, )

DA=CB ( ,, )

FBX=ABE (vertically opposite )

ABE=CBF

FCB=BAE (exterior angle of

parallelogram )

so,ABE similar to CBF

so, DE=DF

and DE=AB+BC

DE=DF=AB+BC hence proved