Question

ABCD is a parallelogram and it's side AD increases to point E and BE line segment cuts DC at F then you have to proove that triangle ABE is similar to Triangle BFC

Answer 1

hey
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given: ABCD is a || gm where AD is produced to E and BE is cuts CD at F

to proove : ∆ABE is congruent to ∆BFC

proof : in ∆ABE and ∆BFC

AD = BC ( opposite sides of ||gm are equal )

then , AE = BC

<A = <C ( opposite angles of ||gm are equal )

as we know that opposite sides of a || gm are parallel

AD ||BC

then , DE ||BC and BE is the transversal

< 1 = <2 ( by alternate interior angles )

hence, ∆ABE is similar to ∆BFC by ASA criteria

for the figure, refer to the attachment

hope helped
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