# Van Khea's Quickie

### Solution

$AI\,$ is the bisector of $\angle BAD,\,$ $AJ\,$ is the bisector of $\angle CAD.\,$ Since $\angle BAC=60^{\circ},\,$ $\angle IAJ=30^{\circ}.\,$

But $\angle IAJ\,$ is inscribed into circle $(K),\,$ implying that it is subtended by the arc $\overset{\frown}{AJ}=60^{\circ}.\,$ The corresponding central $\angle IKJ=60^{\circ},\,$ and, since $\Delta IKJ\,$ is isosceles $(IK=JK),\,$ it is also equilateral.

### Acknowledgment

The above problem was posted by Van Khea (Cambodia) at the Olimpiada pe Şcoală (The School Yard Olympiad) and commented with the above solution by Marian Dincă (Romania).

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