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In the diagram below, AD is perpendicular to AC and $ ∠BAD = ∠DAE = 12^\circ$. If $AB + AE = BC$, find $∠ABC$. Blockquote

The above is the diagram.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it.

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closed as off-topic by heropup, user122283, colormegone, Davide Giraudo, user7530 Jun 14 '14 at 16:45

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  • $\begingroup$ I do not actually know how to do the question on first sight. I am only able to find out the angles of $∠BAD$ , $∠DAE$ and $∠EAC$ but after that, I am not sure how to solve it. $\endgroup$ – snivysteel Jun 15 '14 at 2:33
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Proof.

Take $F$ on $\overrightarrow{BA}$ beyond $A$ such that $BF=BC$
$AE=AF$ and $\triangle AEF$ is isosceles.
$2\angle BAD=\angle BAE=\angle AEF+\angle AFE=2\angle AFE$
Thus $AD$ is parallel to $EF$.
$AD\bot AC$ and then $AC\bot EF$, which tell us $\square AECF$ is kite.
$\angle EFC=90^{\circ}-\angle FCA$, $\angle AFC=\angle AFE+\angle CFE=102^{\circ}-\angle FCA$
$\angle AFC=\angle ECF=2\angle FCA$, since $\triangle BCF$ is isosceles.
Hence, $\angle FCA=34^{\circ}$ and $\angle ABC=44^{\circ}$

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