Question

E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F.Prove that DBE~ ABC

Answer 1

Solution: Since Δ ABE ≈ Δ ACD

Hence, BE = CD

And ∠DBE = ∠ECD

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In Δ DBE and Δ ECD

BE = CD (proved earlier)

∠DBE = ∠ECD (proved earlier)

DE = DE (common)

Hence, Δ DBE ≈ Δ ECD

This means; DB = EC

This also means;

Hence; DE || BC

Thus, Δ ADE ~ Δ ABC proved.