In triangle $ABC$, $AD$ is the angle bisector of angle $A$, and $D$ lies on $BC$. We need to prove that $\frac{AB}{AC} = \frac{DB}{DC}$.

1. Problem Description

In triangle $ABC$, $AD$ is the angle bisector of angle $A$, and $D$ lies on $BC$. We need to prove that $\frac{AB}{AC} = \frac{DB}{DC}$.

2. Solution Steps

Since $AD$ is the angle bisector of angle $A$, by the Angle Bisector Theorem, we have that the ratio of the sides of the triangle is equal to the ratio of the segments created by the angle bisector on the opposite side. Specifically,

$\frac{AB}{AC} = \frac{DB}{DC}$

This is the Angle Bisector Theorem.

Let us provide a proof of the Angle Bisector Theorem.

Let ABC be a triangle and let AD be the angle bisector of angle A, with D lying on BC.

We wish to prove that $\frac{AB}{AC}=\frac{BD}{CD}$.

Draw a line through C parallel to AD, and extend AB to meet this line at E.

Since AD is parallel to CE, angle DAC is equal to angle ACE, and angle BAD is equal to angle AEC.

Since AD bisects angle A, angle BAD = angle DAC.

Therefore, angle AEC = angle ACE.

Hence, triangle ACE is isosceles, with $AE = AC$.

Since AD is parallel to CE, by similar triangles, we have

$\frac{BD}{CD} = \frac{BA}{AE}$

Since $AE = AC$, we have

$\frac{BD}{CD} = \frac{BA}{AC}$

$\frac{AB}{AC} = \frac{BD}{CD}$

$\frac{AB}{AC} = \frac{DB}{DC}$

3. Final Answer

$\frac{AB}{AC} = \frac{DB}{DC}$