Question

E is a point on the side AD produced of a parallelogramABCD and BE intersects CD at F. Show that triangle ABE is similar to triangle CFB​

Answer 1

Given,

ABCD is parallelogram.

To Prove,

ABE ~ CFB

Solution,

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles)

ΔABE ~ ΔCFB (By AA similarity criterion)

Hence Proved.

Extra Information :-

AA similarity criterion - If two angles of one triangle are equal to two angles of another triangles, then the two triangles are similar.