How to prove AC = BD? Triangle ABC has $\widehat{A} = 130^\circ;\, \widehat{B} = 30^\circ.$ D and E are the intersection between the mid-perpendicular of $AC$ and $BC, AB,$ respectively.
a) Prove ADE is an isosceles triangle
b) AC = BD
a) is an easy one, but b) is in a different level (at least with me)
How can I prove AC= BD?
Please help me!

 A: Do you see why $\triangle ADF \cong \triangle DAH$?
That leads to $DF = AH = \frac{AC}{2}$
$BD = 2 \times DF$ (as $ \, \angle FBD = 30^0$)

A: 
Hint:
Draw perpendicular bisector of BD( a perpendicular passing the midpoint M of DB) . Draw a line from D parallel with AC to intersect the drawn bisector at F.We have:
$\angle ADB=\angle CAD=20^o$
$\angle ADB=\angle FDB=\angle FBD=20^o$
Since $\angle ADF=20^o$ F must be on circle center on D and passing A, therefore $AD=DF$. Also $AD=DC$ and $DF=FB$
Hence triangles ACD and FDB for SAS (two sides and angles they make are equal)are equal so AC=BD.
A: Define $D'$ as the reflection of point $D$ over $EB$.
Here's the rough sketch ( only angle chase is required ) :  show $D'CEB$ cyclic $\rightarrow \Delta AD'B_1$ isosceles $\rightarrow CD=AB_1 \rightarrow \Delta CB_1B$ isosceles
Here's the detailed one.


*

*Claim: $D' \in (CEB)$
The proof is only angle chase.
Note that $\angle DAC=\angle DCA=20 \implies\angle DAE=\angle ADE=70 \implies \angle DEA= 40 \implies\angle EDB=110$ .
Also note that $\Delta ECA$ is isosceles, so $\angle CEA =2\cdot \angle DEA\implies CEA =80 \implies ECB=70$
Now since $D'$ is the reflection of point $D$ over $EB$. So $\angle ED'B=\angle EDB=110$
By cyclic quads angles property we get $D'\in (CEB)$.

Now define $CA\cap (ECB)=B_1$

*

*Claim: $AD'B_1$ is isosceles. Hence $CD=AB_1$.

Again since $D'$ is the reflection of point $D$ over $EB$. So $\angle D'EA=\angle AED=\angle DEC=40\implies \angle D'EC=120\implies CB_1D'=60$( using cyclic quad $CB_1D'E$)
Also note that, since $\Delta ECA$ is isosceles and $\angle CEA=80 \implies \angle ACE=50$
So by cyclic quad $CED'B_1$, we get $\angle ED'B_1=130$ and also $\angle ADE=\angle AD'E=70 \implies \angle AD'B_1=\angle ED'B_1-\angle ED'A=60$.
So $AD'B_1$ is isosceles. Also $AD'=AD$ ( $D'$ is reflected point) , So $AD=AB_1$, but $AD=CD\implies CD=AB_1$ .

*

*Claim:$\Delta CB_1B$ isosceles

Consider cyclic quad $ED'B_1B$. We get $\angle D'EB+\angle D'B_1B=180 \implies \angle CB_1B=80$ . But $\angle ACB=20$. So $\Delta CB_1B$ isosceles  .

*

*Claim:$AC=BD$
So we have $CB_1=CB$ , but we also have $CD=AB_1 \implies AC=BD$
And we are done!
