E is the point on side AD produced of a parallelogram ABCD and BE intersect CD at F prove that ABC = CFB
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Explanation:
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.
Solution:
[Fig is in the attachment]
In AABE and ACFB,
ZA = 2C (Opposite angles of a parallelogram)
ZAEB = 2CBF (Alternate interior angles
as AE || BC)
AABE ACFB (By AA similarity criterion)
I hope this may help you
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