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Question
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB.
Solution 1
In ΔABE and ΔCFB,
∠BAE = ∠FCB ...(Opposite angles of a parallelogram)
∠AEB = ∠CBF ...(Alternate interior angles as AE || BC)
∴ ΔABE ∼ ΔCFB ...(By AA similarity criterion)
Solution 2
Given: A parallelogram ABCD where E is a point on side AD produced & BE intersects CD at F.
To Prove: ΔABE ~ ΔCFB.
Proof: In parallelogram ABCD,
opposite angles are equal,
Hence, ∠A = ∠C ...(1)
Also, In parallelogram ABCD opposite sides are parallel, AD || BC
Now since AE is AD extended,
AE || BC
and BE is the traversal
∴ ∠AEB = ∠CBF ...(Alternate Angles) ...(2)
Now in Δ ABE & Δ CFB
∠A = ∠C ...[From (1)]
∠AEB = ∠CBF ...[From (2)]
∴ ΔABE ~ ΔCFB ...(AA similarity criterion)
Hence proved.
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