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BdMO regional 2018 set 4 Secondary P 06
Posted: Thu Jan 17, 2019 2:03 pm
by samiul_samin
$ABCD$ is a quadrilateral, where $BC=CD=2$ and $\angle DCA=\angle DBA$,
$\angle
BAC=60^{\circ}$ . What is the length of $BD$?
Re: BdMO regional 2018 set 4 Secondary P 06
Posted: Thu Jan 17, 2019 5:32 pm
by Tasnood
$\angle DCA=\angle DBA$
So, $ABCD$ is cyclic quadrilateral.
$\angle BAC=\angle BDC=60^{\circ}$
So, $\triangle BDC$ is isosceles and $BD=2$
Re: BdMO regional 2018 set 4 Secondary P 06
Posted: Thu Jan 17, 2019 8:13 pm
by samiul_samin
Tasnood wrote: ↑Thu Jan 17, 2019 5:32 pm
$\angle DCA=\angle DBA$
So, $ABCD$ is cyclic quadrilateral.
$\angle BAC=\angle BDC=60^{\circ}$
So, $\triangle BDC$ is isosceles and $BD=2$
It was more than easy
