Question

Prove that the angle bisector of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Answer 1

Proof : Line $CE\parallel$ Line $DA$ (Construction) and line $BE$ is the transversal
$\angle BAD=\angle AEC$ [Corresponding angles] $....\left(1\right)$
$AD\parallel EC$
$△AC$ is transversal
$\therefore \angle DAC=\angle ACE$ [Alternative angles] $...\left(2\right)$
But $\angle BAD=\angle DAC$ [Given] $...\left(3\right)$
$△\angle AEC=\angle ACE$ [from $\left(1\right),\left(2\right)$ and $\left(3\right)]$
In $△AECsideAC=sideAE$ [Isosceles theorem] $...\left(4\right)$$
Now in $△BCE,segAD\parallel segCE$
$\frac{BD}{DC}=\frac{AB}{AE}$
$\therefore \frac{BD}{DC}=\frac{AB}{AC}$
Hence in a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides.