Question

In ∆ABC, ray AD bisects ∠A and intersects BC in D. If BC = a, AC = b and AC = c, prove that \[BD = \frac{ac}{b + c}\]

Answer 1

Given: In Δ ABC ray AD bisects angle A and intersects BC in D, If `BC = a, AC=b` and  `AB =c` 

To Prove:

\[BD = \frac{ac}{b + c}\]

The corresponding figure is as follows

Proof: In triangle ABC, AD is the bisector of  `∠ A`

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

Substitute `BC = a, AC= b` and `AB =c` we get,

`c/b=(BD)/(BC-BD)`

`c/b=(BD)/(a-BD)`

By cross multiplication we get.

`c(a-BD)=bxxBD`

`ac-cBD=bBD`

`ac=bBD+cBD`

`ac=(b+2)BD`

`(ac)/(b+c)=BD`

We proved that  `BD=(ac)/(b+c)`