Question

In ΔABC, ray BD bisects ∠ABC.

If A – D – C, A – E – B and seg ED || side BC, then prove that:

`("AB")/("BC") = ("AE")/("EB")`

Proof : 

In ΔABC, ray BD bisects ∠ABC.

∴ `("AB")/("BC") = (......)/(......)`   ......(i) (By angle bisector theorem)

In ΔABC, seg DE || side BC

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

∴ `("AB")/square = square/("EB")`   [from (i) and (ii)]

Answer 1

Proof : 

In ΔABC, ray BD bisects ∠ABC.

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

In ΔABC, seg DE || side BC

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

∴ `("AB")/bb(BC) = bb(AE)/("EB")`   ....[from (i) and (ii)]