Question

In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:

Proof :

In ΔABC, ray BD bisects ∠B.

∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

ΔABC, DE || BC

∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]

Answer 1

In ΔABC, ray BD bisects ∠B.

∴ `(bbunderline"AB")/("BC") = ("AD")/("DC")`   ...(I) (By angle bisector theorem)

In ΔABC, DE || BC

∴ `(bbunderline"AE")/("EB") = ("AD")/("DC")`   ...(II) (Basic proportionality theorem)

∴ `("AB")/bbunderline"BC" = bbunderline"AE"/("EB")`   ...[from (I) and (II)]