Interior angle bisector theorem

Submitted by Structure on Thu, 12/20/2007 - 20:33.

Let ABC be a triangle. Let AM be the bisector of the angle A. Then we have:

$\frac{AB}{AC}=\frac{MB}{MC}$

Proof:

Let d be the line parallel at the line AM that passes through C. Let F be the intersection between this line d and AB.

AFC = BAM (because AM is parallel with FC)
BAC = MAC (because AM is the bisector of BAC)
MAC = ACF (because AM is parallel with FC)

Because AFC = ACF the triangle AFC is isoscele. So AF = AC.

From the Thales theorem we have:

$\frac{BM}{MC}=\frac{BA}{AF}$

because AF = AC we have:

$\frac{BM}{MC}=\frac{BA}{AC}$