Originally Posted by

**thereddevils** In the figure , ABCD is a parallelogram . AE=BF and C is parallel to FH . Prove that

(1) EFCD is a parallelogram

(2) GHFC is a parallelogram

(3) parallelogram GHFC and ABCD are equal in area

My work :

(1) BC // AD , AE=BF , AD=BC

$\displaystyle \angle ADE =\angle CBF $(corresponding angle) No. You mean $\displaystyle \color{red}\angle DAE=\angle CBF$

$\displaystyle \triangle ADE$ congruent to $\displaystyle \triangle BCF $

$\displaystyle \angle DEA =\angle CFB $

DE// CF , AF// DC

Hence , EFCD is a paralleogram. Apart from that this is fine.

(2) DH // CF , CG//HF

Hence , GHFC is a parallelogram . OK.

(3) not really sure .Use the result from your previous post: parallelograms on the same base and between the same parallels are equal in area. Look first at **CF**DE and **CF**GH; then at **CD**EF and **CD**AB. (Note the bold type.)