Question

Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Answer 1

Given: A triangle ABC with AD as the internal bisector of angle A, meeting BC at D.

To Prove: `(BD)/(DC) = (AB)/(AC)`. This is the Angle‑Bisector Theorem.

Proof [Step-wise]:

1. Since AD is the internal bisector of ∠A, we have ∠BAD = ∠CAD.   ...(Given)

2. Area(ΔABD) can be written using two sides and the included angle: 

Area(ΔABD) = `1/2` × AB × AD × sin(∠BAD)

3. Similarly, Area(ΔADC) = `1/2` × AC × AD × sin(∠CAD).

4. From step 1, ∠BAD = ∠CAD, so sin(∠BAD) = sin(∠CAD).

With AD common, divide the expressions in steps 2 and 3 to get `(Area(ΔABD))/(Area(ΔADC)) = (AB)/(AC)`.

5. Note that ΔABD and ΔADC share the same altitude from A to the line BC, so area is proportional to the base on BC.

Hence, `(Area(ΔABD))/(Area(ΔADC)) = (BD)/(DC)`.

6. Equating the two expressions for the area ratio (steps 4 and 5) yields `(BD)/(DC) = (AB)/(AC)`.

Therefore, the internal bisector AD of angle A divides the opposite side BC internally in the ratio BD : DC = AB : AC, as required.