Question
In $\triangle\text{ABC},$ ray AD bisects $\angle\text{A}$ and intersects BC in D. If BC = a, AC = b and AB = c, prove that
$\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}$
$\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}$
✓
Answer
Given: In $\triangle\text{ABC}$ ray AD bisects angle A and intersects BC in D,
If BC = a, AC = b and AB = c
$\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}$
The corresponding figure is as follows
Proof: In triangle ABC, AD is the bisector of $\angle\text{A}$
Since BC = CD + BD
⇒ CD = BC - BD
$\text{CD}=\text{a}-\frac{\text{ac}}{\text{b}+\text{c}}$
$=\frac{\text{ab}}{\text{b}+\text{c}}$
If BC = a, AC = b and AB = c
$\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}$
The corresponding figure is as follows
Proof: In triangle ABC, AD is the bisector of $\angle\text{A}$
Since BC = CD + BD
⇒ CD = BC - BD
$\text{CD}=\text{a}-\frac{\text{ac}}{\text{b}+\text{c}}$
$=\frac{\text{ab}}{\text{b}+\text{c}}$
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