E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
In ∆ABE and ∆CFB, we have
∠AEB = ∠CBF [all angles]
![In ∆ABE and ∆CFB, we have∠AEB = ∠CBF [all angles]∠A =](/application/zrc/images/qvar/MAEN10042218.png)
∠A = ∠C [opp. ∠s of a ||gm]
∴ By A.A. criterion of similarity, we have
∆ABF ~ ∆CFB. Hence Proved.
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