Proving that $MA + MB > CA + CB$ in the given figure

Question

On the extension of side $BC$ of the triangle $ABC$ consider the point D such that $C$ is between $B$ and $D$ and $AC = CD.$ Let M be an arbitrary point on the angle bisector of $\angle ACD$. Prove that $MA + MB > CA + CB.$

I am completely stuck on this question. I don't know how to approach it. Any help would be greatly appreciated.

Answer 1

Join $MD$. Here I have assumed that $\triangle MBD$ is not degenerate.

Here's a proof $$\triangle AMC \cong \triangle DMC$$ By $SAS$ congruence criterion. So $$AM=MD$$ Triangle inequality in $\triangle MBD$ gives $$MB+MD>BC+CD$$ which gives using previous equality and the fact $AC=CD$ $$MB+MA>CA+CB$$ $QED$

Answer 2

Connect $M$ to $D$ to form the line $MD$.

In triangles $ACM$ and $CDM$,

1) $\angle ACM = \angle MCD$ (because $CM$ is the angle bisector of $\angle ACD$).

2) The side $CM$ is common.

3) $AC=CD$ (They were chosen that way)

Now, we have that the given triangles are congruent (I think it's called the $SAS$ criteria), so that $AM=MD$.

Now, simply note that $BM+MD > BD$ using triangle inequality, and $BD = BC+CD = BC + AC$, while $DM=AM$, so this gives $BM + AM > BC + AC$.