$ABCD$ 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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- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm

### 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 $BD=2$

- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm