Question

Question 7
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that $\Delta OCD$ is an isosceles triangle.

Answer 1

O is a point in the interior of a square ABCD such that $\Delta OAB$ is an equilateral triangle.

Construction Join OC and OD.
AOB is an equilateral triangle.
$\therefore \angle OAB=\angle OBA={60}^{\circ }...\left(i\right)$
Also, $\angle DAB=\angle CBA={90}^{\circ }...\left(ii\right)$ [each angle of a square is ${90}^{\circ }]$
$[\because$ ABCD is a square]
On subtracting Eq. (i) from Eq. (ii) we get
$\angle DAB-\angle OAB=\angle CBA-\angle OBA={90}^{\circ }-{60}^{\circ }i.e.,\angle DAO=\angle CBO={30}^{\circ }$
In $\Delta AODand\Delta BOC,$
AO = BO [given]
[all the side of an equilateral triangle are equal]
$\angle DAO=\angle CBO$ [proved above]
and AD = BC [sides of a square are equal]
$\therefore \Delta AOD\cong \Delta BOC$ [by SAS congruence rule]
Hence, OD = OC [by CPCT]
In $\Delta COD,$
OC = OD
Hence, $\Delta COD$ is an isosceles triangle.