Question

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

Answer 1

Given that:

$ABCD$ is a square and $△OAB$ is an equilateral triangle.

Therefore,

$AB=BC=CD=DA$

$OA=OB=AB$

To Prove:

$∆OCD$ is an isosceles triangle.

Proof

In square $ABCD$

$\angle 1=\angle 2=90°$

$\angle 1=\angle 2$ --------------$\left(A\right)$

From equilateral $△OAB$

$\angle 3=\angle 4=60°$

$\angle 3=\angle 4$ ---------------$\left(B\right)$

Subtract equation $\left(A\right)$ from equation $\left(B\right)$

$\angle 1-\angle 3=\angle 2-\angle 4$

$\angle 5=\angle 6$ ( From figure it can be seen)

Now,

In $△DOA$ and $△COB$

$AD=BC$ ( Side of square )

$\angle 5=\angle 6$ ( Proved above )

$OA=OB$ ( Side of equilateral )

By $SAS$ criterion of congruence

( $SAS$ congruence rule: If any two sides and the angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by $SAS$ rule )

Since, $△DOA\cong △COB$

Therefore, $OD=OC$

Hence, proved that $△COD$ is an isosceles triangle.