$ABCD$ is a quadrilateral, where $BC=CD=2$ and $\angle DCA=\angle DBA$,
$\angle
BAC=60^{\circ}$ . What is the length of $BD$?
BdMO regional 2018 set 4 Secondary P 06
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Re: BdMO regional 2018 set 4 Secondary P 06
$\angle DCA=\angle DBA$
So, $ABCD$ is cyclic quadrilateral.
$\angle BAC=\angle BDC=60^{\circ}$
So, $\triangle BDC$ is isosceles and $BD=2$
So, $ABCD$ is cyclic quadrilateral.
$\angle BAC=\angle BDC=60^{\circ}$
So, $\triangle BDC$ is isosceles and $BD=2$
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