How to prove this conjecture with Converse problem Problem:Let $\Delta ABC$,such the point $D$ on $AB$,such $$AD=AC,~~~~CD=AB,~~~~\angle BAC=100^{\circ}$$ show that: $\angle ACB=50^{\circ}$
I have solve this problem,Following is my solution:


take the $E$ such $\Delta ABE$ is equilateral,since
  $$\angle BAC=100^{\circ},AD=AC\Longrightarrow \angle EAC=40^{\circ}=\angle ADC$$
  so we have
  $$\Delta ADC\cong\Delta CAE$$ so we have
  $$AC=CE$$
  since
  $\Delta ABE$is equilateral,so we have
  $AB=BE$
  so we have
  $$\Delta ABC\cong EBC$$
  then we have
  $$\angle ABC=\angle EBC=\dfrac{60^{\circ}}{2}$$
  so the $$\angle ACB=50^{\circ}$$

I conjecture the conversen problem is also right.But How prove it
Converse problem: 
Let $\Delta ABC$,such the point $D$ on $AB$,such $$AD=AC,~~~~CD=AB,~~~~\angle ACB=50^{\circ}$$ show that: $\angle BAC=100^{\circ}$

 A: Let's call $\angle ABC =2x$. From the triangle $ACD$ we have
$$\sin x=\frac{DC}{2AC} \quad (1)$$
Sine rule at the triangle $ABC$:
$$\frac{AC}{\sin(50°+2x)}=\frac{AB}{\sin(50°)} \quad(2)$$
We know that $AB=DC$, so putting $(1)$ and $(2)$ together we have:
$$2\cdot \sin x\cdot\sin (50°+2x)=\sin 50°$$
Now, solving the above equation by computer we can see two solutions:
$$x=50° \quad \text{and} \quad x=22,62°$$
So your guess is not true.
We have two possible values for $\angle BAC$ which are $100°$ and $45,24°$.  
A: $\angle ABC=\alpha$ , $0< \alpha < 50^\circ$
Let $R$ be the radius of the circumscribed circle $ABC$.
$$AB = 2R\cdot sin50^\circ$$
$AC= 2Rsin\alpha$
$\angle BAC = 130^\circ - \alpha$
$$CD = 2AC \left( sin\frac{1}{2}\angle { BAC} \right)
 = 4R\cdot sin\alpha \cdot sin\left(65^\circ - 
\frac{\alpha}{2} \right) = f(\alpha)$$
I want to show that $f(\alpha)$ is strictly increasing function.
If it is true, then the number of $\alpha$ satisfying $ AB = CD $ is at most one. 
And we know that there is one existing case. Thus the converse problem is proved.  
$f'(\alpha)= 
4R \cdot cos\alpha \cdot sin\left(65^\circ - 
\frac{\alpha}{2} \right) - 
2R \cdot sin\alpha \cdot cos\left(65^\circ - 
\frac{\alpha}{2} \right) $
When $\alpha$ is increasing, both $cos\alpha$ and 
$ sin\left(65^\circ - \frac{\alpha}{2} \right)$ are decreasing, 
and both $sin\alpha$ and 
$ cos\left(65^\circ - \frac{\alpha}{2} \right) $ are increasing. 
So $f'(\alpha)$ is decreasing function, and 
$$ f'(\alpha) \ge f'(50^\circ) = 
4R \cdot cos50^\circ \cdot sin40^\circ - 
2R \cdot sin50^\circ \cdot cos40^\circ 
\approx 0.479 R > 0$$
Done.
