Provide four proof methods:

As shown in the figure, AM is the angular bisector of △ABC, which proves that AB/AC = MB/MC.

Proof: method 1: (area method)

The area of triangle ABM is S=( 1/2)*AB*AM*sin∠BAM,

Triangle ACM area S=( 1/2)*AC*AM*sin∠CAM,

So triangle ABM area S: triangle ACM area S=AB:AC.

Triangle ABM and triangle ACM are equal-height triangles, and the area ratio is equal to the bottom ratio.

That is, triangle ABM area s: triangle ACM area S=BM:CM.

So ab/AC = MB/MC

Method 2 (Similarity)

Let c be the extension of CN parallel to AB and AM in n.

Triangle anti-missile, similar triangle NCM,

AB/NC=BM/CM，

You can also prove that ∠CAN=∠ANC.

So AC=CN,

So ab/AC = MB/MC

Method 3 (Similarity)

Crossing m makes MN parallel to AB and AC in n.

Triangle ABC similar triangles NMC,

AB/AC=MN/NC，AN/NC=BM/MC

It can also be proved that ∠CAM=∠AMN.

So AN=MN,

So AB/AC=AN/NC

So ab/AC = MB/MC

Method 4 (Sine Theorem)

Draw the circumscribed circle of a triangle and the intersection of AM and d,

According to sine theorem,

AB/sin∠BMA=BM/sin∠BAM，

AC/sin∠CMA=CM/sin∠CAM

∠BAM=∠CAM，∠BMA+∠AMC= 180。

sin∠BAM=sin∠CAM，sin∠BMA=sin∠AMC，

So ab/AC = MB/MC