Question

State the basic proportionality theorem. Use the theorem to prove the following:

In ΔABC, AD is the angle bisector of angle A. BA is produced to E such that CE || AD. Prove that `(BD)/(DC) = (BA)/(AC)`.

Answer 1

Given:

In ΔABC, AD is the internal bisector of ∠A.

BA is produced to E (so B, A, E are collinear) and CE || AD.

To Prove: `(BD)/(DC) = (BA)/(AC)`

Proof (Step-wise):

1. Consider ΔBCE. The line AD meets BE at A and BC at D and AD || CE by hypothesis.

Hence, by the Basic Proportionality Theorem applied to ΔBCE, we get `(BA)/(AE) = (BD)/(DC)`.

2. Show AE = AC.

Since CE || AD, ∠ACE (angle between AC and CE) equals ∠CAD (alternate interior angles).

Also CE || AD and AE is the extension of AB, so ∠AEC (angle between AE and EC) equals ∠BAD (alternate interior angles).

But AD is the angle bisector, so ∠BAD = ∠CAD.

Therefore ∠AEC = ∠ACE, so ΔAEC is isosceles and hence AE = AC.

3. Substitute AE = AC into the relation from step 1:

`(BD)/(DC) = (BA)/(AE) = (BA)/(AC)`.

Thus `(BD)/(DC) = (BA)/(AC)`, as required.