Angle Bisector Problems

Question

In a triangle $ABC$, the bisector $AD$ and $BE$ intersect each other at $X$. Write the comparison of $AX:XD$ by considering $a, b, c$ as the lengths of the side of that triangle.

I don't know how to continue after drawing the triangle and its properties given. Should I use trigonometry or anything else?? Maybe such similarity and congruence theorem works, but could you give me ideas?

Answer 1

Hint

First of all, use Menelaus's Theorem at the triangle $ADC$ and secant $BXE$

$$\frac{BC}{BD}\cdot \frac{XD}{AX}\cdot \frac{EA}{EC}=1$$

Now use bisector's theorem at the triangle $ABC$

$$\frac{BD}{c}=\frac{CD}{b}\to \frac{BD}{c}=\frac{BC-BD}{b}\to \frac{BC}{BD}=\frac{c}{c+b}$$

Use bisector theorem again to get $\frac{EA}{EC}$.

Can you finish?

Answer 2

With the Theorem sines we get $$AX=\frac{c\sin\left(\frac{\beta}{2}\right)}{\sin\left(\frac{\beta+\gamma}{2}\right)}$$ $$XD=\frac{\sin\left(\frac{\beta}{2}\right)\cdot BD}{\sin\left(\frac{\beta+\gamma}{2}\right)}$$ $$BD=\frac{ac}{b+c}$$ so we get $$\frac{AX}{BX}=\frac{b+c}{a}$$