BdMO regional 2018 set 4 Secondary P 06

Forum rules
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.
samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
BdMO regional 2018 set 4 Secondary P 06

Unread post by samiul_samin » Thu Jan 17, 2019 2:03 pm

$ABCD$ is a quadrilateral, where $BC=CD=2$ and $\angle DCA=\angle DBA$,
$\angle
BAC=60^{\circ}$ . What is the length of $BD$?

User avatar
Tasnood
Posts:73
Joined:Tue Jan 06, 2015 1:46 pm

Re: BdMO regional 2018 set 4 Secondary P 06

Unread post by Tasnood » 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 $BD=2$ :D

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

Re: BdMO regional 2018 set 4 Secondary P 06

Unread post by samiul_samin » Thu Jan 17, 2019 8:13 pm

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 $BD=2$ :D
It was more than easy :P :P :P

Post Reply