1. ## Parallelogram (2)

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 :

$\angle ADE =\angle CBF$(corresponding angle)

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

$\angle DEA =\angle CFB$

DE// CF , AF// DC

Hence , EFCD is a paralleogram

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

Hence , GHFC is a parallelogram .

(3) not really sure .

2. Hello thereddevils

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 :

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

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

$\angle DEA =\angle CFB$

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 CFDE and CFGH; then at CDEF and CDAB. (Note the bold type.)