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!


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$)

