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For reference: Calculate $x~ if AE = BC, and AD = BE$ enter image description here

My progress: I tried for auxiliary lines...parallel to BC by A and parallel to BE by A... forming the parallelograms...I completed the Angles but I was unsuccessful...but I believe the solution is in the auxiliary lines

enter image description here

For illustration...DR SK's resolution enter image description here

peta arantes
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    This is similar to this question of yours: https://math.stackexchange.com/q/4257751/947379 The only difference is you have to apply that method twice, like MOBINULS's answer does. – ACB Sep 28 '21 at 14:29
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    @ACB ...well remembered! – peta arantes Sep 28 '21 at 14:55

1 Answers1

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Let the circumcentre of $\triangle ABC$ be $O$. $\triangle OBC$ is equilateral.

$\Rightarrow AE=BC=AO$

$\triangle AOE$ is isosceles. Through angle chasing, $\angle BOE=\angle COE=30^{\circ}$ and thereafter $BE=EC$ and $\angle BEA=40^{\circ}$.

Let the circumcentre of $\triangle ABE$ be $P$. Again, $\triangle PBE$ is equilateral, and through angle chasing, $PB$ passes through $D$. Hence, $x=60^{\circ}$.

Limestone
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