Question

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABC is similar to ΔCFB.

Answer 1

Answer:

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Answer 2

Answer:

In a parallelogram opposite angles are equal to each other. Then find the relation between angles of both triangle's to show them similar.

If two angles of a triangle are respectively equal to two angles of another triangle, then Triangles are similar (because by the angle sum property of a triangle their third angle will also be equal ) and it is called AA similarity criterion.

In ΔABE and ΔCFB,

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

∠AEB=∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ΔCFB (By AA similarity criterion)

solution