Question:
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Updated On: Jan 13, 2026
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Solution and Explanation
Given: ABCD is a parallelogram, where BE intersects CD at F. E is a point on the side AD produced.
To Prove: ΔABE ~ ΔCFB
Proof: In ∆ABE and ∆CFB,
\(\angle\)A = \(\angle\)C (Opposite angles of a parallelogram)
\(\angle\)AEB = \(\angle\)CBF (Alternate interior angles as AE || BC)
∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)
Hence Proved
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