MathTime

... Or construct the isosceles triangle ADO with $\widehat{AOD}=100^\circ$, C and O on the same side of AB; as triangles ADO and ABC are congruent (s.a.s.) it follows that triangle AOC is equilateral wherefrom O is the circumcenter of triangle ADC and $2m(\widehat{ADC})=m(\widehat{AOC})=60^\circ$. B...

Another nice idea: Construct the equilateral triangle $\triangle ACO$, $O,B$ on the same side of $AC$, then $\triangle OAD\equiv\triangle ABC$ (s.a.s.), so $\angle AOD=100^\circ$ and $OD=OC$, therefore $O$ is the circumcenter of $\triangle ADC$, wherefrom $\angle ODC=10^\circ$ and consequently $\ang...

Take ${F}\in BE\cap CD$ and $G$ reflection of $F$ in $BC$; $G$ lies onto $AB$. Now take ${H}\in GF\cap AC$. Since $\angle DGH=\angle DCH=10^\circ$ we get $GDHC$ cyclic hence, simple angle inspection shows $DHEF$ cyclic as well, with $\angle EDH=10^\circ$, finally giving $\angle EDF=60^\circ\iff\angl...

Take $O$ the circumcenter of $\triangle BCE$; it belongs to $CD$. Because $\angle BEO=\angle BDO=60^\circ$, $BDEO$ is cyclic, thus $\angle CDE=\angle ODE=60^\circ$, making $\angle ADE=60^\circ$.

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sunken rock

Let the perpendicular bisector of $AC$ be cut by $CE,CB$ at $F,G$ respectively. Then $AFBC$ is cyclic and $\measuredangle EFB=20^\circ\ (\ 1\ )$. Since the reflection of $D$ across $AB$ belongs both to $AG$ and $BC$, we conclude that $G$ is that reflection. But $GFEB$ is cyclic from $\angle GBE+\ang...

Remark: the problem can be generalized, as follows:
If $m(\widehat{ABC})=\alpha> 45^\circ$ and $m(\widehat{ACE})=2\alpha-90^\circ$, $CE=AB$, gives $\triangle ABD$ isosceles.

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sunken rock

$D$ lies onto $AB$!

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sunken rock

Hint: $C$ is the circumcenter of $\triangle AEF$.

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sunken rock

Hint: Construct the equilateral triangle $ABE, AB$ separating $C$ and $E$, see that $C-D-E$ are collinear. According to van Schooten theorem $CE=AC+CB\ (\ 1 \ )$, but $CD\cdot CE=CA\cdot CB$; with $(1)$ we are done.

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sunken rock

Hint: After some analysis the problem is equivalent with: On the smaller arc $BC$ of the circumcircle of the equilateral triangle $ABC$ take a point $S$. The perpendicular bisector of $BS$ intersects $AC$ at $P$ and the perpendicular bisector of $CS$ intersects $AB$ at $Q$, then $PQ$ is tangent to t...