Question

in the given figure , O is a point in the interior of square ABCD such that triangle OAB is an equilateral triangle.... show that triangle OCD is an isosceles triangle​

Answer 1

Solution :

Given that Triangle OAB is an equilateral triangle so we can write

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

Also given that ABCD is a square so we can write

$\sf \angle \: A = \angle \: B = \angle \: C = \angle \: D = {90}^{ \circ}$

To find the value of angle DAO, we can write

$\sf \angle \: A = \angle \: DAO + \angle \: OAB$

Put the value of angle A = 90° and OAB = 60°

$\longrightarrow \sf { \: 90}^{ \circ} = \angle \: DAO + {60}^{ \circ}$

$\longrightarrow \sf \angle \: DAO = {90}^{ \circ} - {60}^{ \circ}$

$\longrightarrow \sf \angle \: DAO = {30}^{ \circ}$

When DAO = 30°, CBO too would be 30°

In Triangle OAD and OBC

• AD = BC (Side of the square)

• OA = BC (Side of the equilateral triangle)

• Angle DAO = Angle CBO

By SAS congruence criterion :

$\sf \triangle \: OAD \cong \triangle \: OBC$

So,

OD = OC (By C.P.C.T)

$\therefore$ Triangle OCD is an isosceles triangle.

Proved!

Answer 2

Answer:

OCD is isolaslice the correct answer