Prove that $AD+CD=AB$ In $\triangle ABC$, $AC=BC$, $\angle ACB=100^{\circ}$. $D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$. Prove that $AD+CD=AB$.


This post serves a purpose of recording my solution to this geometry problem, without trigonometry. This isn't a question waiting to be answered.
Feel free to comment or post you thoughts or, if any, alternative answers. If you've spotted mistakes in my answer, you can point them out as well.
 A: 
Let $L$ be a point on $AB$ such that $AD=AL$ and P be the reflection of point $L$ over $AD$.
By simple angle chasing you can get $\angle DPC=80^o$ and $\angle DCP = 80^o$ then you can find relationship $CD=DP$. Also in $\triangle DLB$ you can get $\angle LDB=40^o$ and $\angle LBD=40^o$ and $DL=LB$
And also by properties of a reflection, you can find out that $DL=DP=CD$
Then finally you can get,
$\boxed{AB=AD+CD}$
P.S:- As $AD$ is the bisector of $\angle BAC$, DCP will be a straight line
A: 

Construct $AE=AB$ such that $D$ is on $AE$. With $\angle EAB=20^{\circ}$ we immediately have $\angle ABE=\angle AEB=80^{\circ}$, and thus $\angle DBE=40^{\circ}$. This implies that $BD$ bisects $\angle ABE$.
Extend $AC$ and $BE$. Let $F$ be their intersection. It's clear that $D$ is the incenter of $\triangle ABF$, and we obtain $\angle BFD=30^{\circ}$.
Notice that $\angle BCF=180^{\circ}-\angle ACB=80^{\circ}$. Hence $\angle BCF=\angle BED$, and by AA similarity, $$\triangle BED\sim\triangle BCF,$$ implying $$\frac{BE}{BC}=\frac{BD}{BF}\implies\frac{BE}{BD}=\frac{BC}{BF}.$$ By SAS similarity we can say $$\triangle BEC\sim\triangle BDF,$$ yielding $\angle BCE=\angle BFD=30^{\circ}$. Note that $\angle CDE=\angle ADB=120^{\circ}$, so $\angle CED=30^{\circ}$, which implies $CD=DE$. Finally, $$AB=AE=AD+DE=AD+CD.$$
A: Let $AC=L$, since $\triangle ABC$ is isosceles, we have $AB=2L\cos(40^\circ)$. Next, use sine law:
$$\frac{AC}{\sin(60^\circ)}=\frac{CD}{\sin(20^\circ)}=\frac{AD}{\sin(100^\circ)}\Rightarrow CD=\frac{\sin(20^\circ)}{\sin(60^\circ)}L,~~AD=\frac{\sin(100^\circ)}{\sin(60^\circ)}L$$
We get: $$AD+CD=\frac{\sin(100^\circ)+\sin(20^\circ)}{\sin(60^\circ)}L=\frac{2\sin(60^\circ)\cos(40^\circ)}{\sin(60^\circ)}L=2L\cos(40^\circ)=AB$$
