Parallelogram (2)

thereddevils · #1 $Old$ November 18th, 2009, 06:37 AM

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

$\angle ADE =\angle CBF$ (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 .

Grandad · #2 $Old$ November 18th, 2009, 07:14 AM

Hello thereddevils

Comments below.

Quote:

Originally Posted by thereddevils $View Post$

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

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

Grandad