Find the angle ADE of the given triangle.

Given :
Angle ABE = 50, Angle DAC = 20, Angle BAD = 60, Angle ACB = 20 and Side CD = Side DA.

Unit of angle values given is degree.

Find : Angle ADE.

I'm stuck and I need help here.

Solution 1:

This problem keeps reappearing. Probably this is the $n$-th duplicate.

The information $CD=DA$ is redundant. It follows from $\angle \, DAC = \angle \, ACB = 20^{\circ}$.

Hints:

What is $\angle \, AEB$? Then focus on triangle $ABE$. Anything special about it? Then choose point $F$ on $CA$ so that $\angle \, ABF = 60^{\circ}$. Let $G$ be the intersection point of $AD$ and $BF$. What can you say about triangles $$ABG, \,\, AEG, \,\, EFG, \,\, DFG?$$ What is special about them? Now, is the role of $ED$ clearer? Finally, how much is $\angle \, ADE \,$?