Question 46
Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\angle ADC = 150^\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$.
Prove that if $\angle BAE + \angle CBE = 60^\circ$, then $\angle BDC + \angle EAC = 90^\circ$.
Prove that if $\angle BAE + \angle CBE = 60^\circ$, then $\angle BDC + \angle EAC = 90^\circ$.