Does anybody have solutions?

$\widehat A \equiv \measuredangle CAB \equiv \measuredangle CAD = 50^\circ$ and $\measuredangle CDE = 50^\circ.$ Assume $\measuredangle ADE = 50^\circ$ as well $\iff$ $\measuredangle ADC = 100^\circ$ and $\measuredangle OCA \equiv \measuredangle DCA = 180^\circ - (\measuredangle CAD + \measuredangle ADC) = 30^\circ.$ From O-isosceles $\triangle AOC$ $\implies$ $\measuredangle AOC = 120^\circ$ $\iff$ $\widehat B \equiv \measuredangle ABC = \tfrac{1}{2} \measuredangle AOC = 60^\circ$ and $\widehat C \equiv \measuredangle BCA = 180^\circ - (\measuredangle CAB + \measuredangle ABC) = 70^\circ.$ Conversely, starting from $\triangle ABC$ with $\widehat A = 50^\circ,$ $\widehat B = 60^\circ$ and $\widehat C = 70^\circ$ $\implies$ $\measuredangle AOC = 2 \measuredangle ABC = 120^\circ$. From O-isosceles $\triangle AOC$ $\implies$ $\measuredangle DCA \equiv \measuredangle OCA = 30^\circ$ and $\measuredangle ADC = 180^\circ - (\measuredangle DCA + \measuredangle CAD) = 100^\circ.$ In addition, $-\frac{\overline{EA}}{\overline{EC}} = \frac{\sin (2 \widehat C)}{\sin (2 \widehat A)} = \frac{\sin 140^\circ}{\sin 100^\circ} = \frac{\sin 40^\circ}{\sin 80^\circ} = \frac{1}{2 \cos 40^\circ} =$ $\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 30^\circ}{\sin 50^\circ} = \frac{\sin \widehat {DCA}}{\sin \widehat {CAD}} = \frac{[DA]}{[DC]}$ $\iff$ $DE$ bisects $\measuredangle ADC = 100^\circ.$ As a result, $\measuredangle CDE = 50^\circ,$ satisfying problem condition, and $\measuredangle ADE = 50^\circ$ as well.

See the JPG file. Sorry, I have replaced the old JPG file with a newer one that has a little simpler solution.

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