Question

In the given figure, $O$ is a point in the interior of $1M$ a square $ABCD$ such that $OAB$ is an equilateral triangle. Show that $OCD$ is an isosceles triangle.

Answer 1

We know that $△OAB$ is an equilateral triangle
So it can be written as
$\angle OAB=\angle OBA=AOB={60}^{\circ }$
From the figure we know that $ABCD$ is a square
So we get
$\angle A=\angle B=\angle C=\angle D={90}^{\circ }$
In order to find the value of $\angle DAO$
We can write it as
$\angle A=\angle DAO+\angle OAB$
By substituting the values we get
${90}^{\circ }=\angle DAO+{60}^{\circ }$
On further calculation
$\angle DAO={90}^{\circ }-{60}^{\circ }$
By subtraction
$\angle DAO={30}^{\circ }$
We also know that $\angle CBO={30}^{\circ }$
Considering the $△OAD$ and $△OBC$
We know that the sides of a square are equal
$AD=BC$
We know that the sides of an equilateral triangle are equal
$OA=OB$
By $SAS$ congruence criterion
$△OAD\cong △OBC$
So we get $OD=OC(c.p.c.t)$
Therefore, it is proved that $△OCD$ is an isosceles triangle.