Question

$O$ is a point in the interior of a square $ABCD$ such that $OAB$ is an equilateral triangle show that $△OCD$ is an isosceles triangle

Answer 1

Given, $\Delta OAB$ is an equilateral triangle,

$\angle AOB=\angle OAB=\angle OBA={60}^{\circ }$

$OA=AB=OB$

Also it's given, $ABCD$ is a Square

$\angle DCB=\angle BAD=\angle ABD=\angle CDA={90}^{\circ }$ ------ (ii)

$DA=AB=CB=CD$

Consider,

$\angle OAB=\angle OBA={60}^{\circ }$ ------ (i)

$\angle BAD=\angle ABD={90}^{\circ }$ ------ (ii)

Subtracting (i) from (ii), we get,

$\angle BAD-\angle OAB=\angle ABD-\angle OBA={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

i.e., $\angle DAO=\angle CBO={30}^{\circ }$ ------ (iii)

Now, in $△AOD$ and $△BOC$

$AO=BO$ [$\Delta OAB$ is an equilateral triangle]

$\angle DAO=\angle CBO={30}^{\circ }$ [from (iii)]

$AD=BC$ [$ABCD$ is a square]

Hence, by $SA$ congruence criterion,

$△AOD\cong △BOC$

Hence, $DO=OC$

$\therefore \Delta OCD$ is an isosceles triangle. $\left[Henceproved\right]$