[Figure: Δ ABC with D on side AC (A – D – C); ray BD bisects ∠ABC; E is on side AB (A – E – B) and seg DE || side BC.]In Δ ABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), comple…

Q.3 (A)(1)

[Figure: Δ ABC with D on side AC (A – D – C); ray BD bisects ∠ABC; E is on side AB (A – E – B) and seg DE || side BC.]

In Δ ABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity :

Proof :

In Δ ABC, ray BD bisects ∠B

∴ \(\dfrac{____}{BC} = \dfrac{AD}{DC}\) ............ (I) (____)

In Δ ABC, DE || BC

∴ \(\dfrac{____}{EB} = \dfrac{AD}{DC}\) ............ (II) (____)

\(\dfrac{AB}{____} = \dfrac{____}{EB}\) ............ [from (I) and (II)]