E is point on the side AD produced of a parallelogram ABCD and BE intersects CD at E. Show that \(\triangle\) ABE ~ \(\triangle\) CFB

Solution :

In triangles ABE and CFB, we have,

\(\angle\) AEB = \(\angle\) CBF           (alternate \(\angle\)s are equal)

\(\angle\) A = \(\angle\) C                      (opposite angles of parallelogram are equal)

\(\therefore\) By AA similarity,

\(\triangle\) ABE ~ \(\triangle\) CFB

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