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Question
In ∆ABC, ray AD bisects ∠A and intersects BC in D. If BC = a, AC = b and AC = c, prove that \[BD = \frac{ac}{b + c}\]
Solution
Given: In Δ ABC ray AD bisects angle A and intersects BC in D, If `BC = a, AC=b` and `AB =c`
To Prove:
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)`
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