Question

In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

\[\frac{BD}{CD} = \frac{AB}{AC}\]

Answer 1

Given:

• In ∆ABC, AD is the bisector of the exterior ∠A.
• AD meets the extended side BC at D.
• We need to prove that: 
  `(BD)/(CD)=(AB)/(AC)`

Since AD is the bisector of ∠A’s exterior angle, it creates two equal angles: ∠BAD = ∠CAD

Also, from the given figure, we can see that: ∠ABD = ∠ACD

Proving the Triangles are Similar

From ∆ABD and ∆ACD, we can observe that:

• ∠BAD = ∠CAD (Given, since AD is an exterior angle bisector)
• ∠ABD = ∠ACD (Vertically opposite angles)

Since ∆ABD ∼ ∆ACD, we know that the corresponding sides of similar triangles are in the same ratio: `(BD)/(CD) = (AB)/(AC)`

Since we have proved that:

`(BD)/(CD) = (AB)/(AC)`

this means that the exterior angle bisector divides the extended side in the same ratio as the other two sides of the triangle.