Given a point $x$ inside the triangle $ΔABC$, prove that $XA+XB < CA + CB$

Question

Given a point $x$ inside the triangle $ΔABC$, prove that $XA+XB < CA + CB$

I have used triangle inequality in every way possible, nonetheless, I haven't come up with a proof, since I always get inequalities like

$CA + BA + XA + XB > 2AB$

or

$ XA+XB+2XC > AC+BC $

I have a feeling that this is one of those theorems of euclidean geometry in which there is a key-difficult-to-see step, and after finding it out, the rest comes easy. However, I haven't been able to crack it.

Could you help me with this one?

Thanks in advance.

Answer 1

Extend $AX$ to meet $BC$ at $M$. Applying triangle inequality in $\Delta CAM$ and $\Delta BXM$, we get, $$CA+CM>AX+XM$$ $$BM+XM>BX$$

Add the inequations, and get the result.

Hope it helps:)