DEC is an equilateral triangle in a square ABCD. If BD and CE intersect at O and ∠COD = x°, find the value of x.

Given,

ABCD is a square,

∠ADC = 90°

The diagonal BD bisects the ∠ADC at the vertex.

∠BDC = $\dfrac{∠ADC}{2} = \dfrac{90°}{2}$ = 45°

DEC is an equilateral triangle.

∠DCE = ∠CDE = 60°

From figure,

∠OCD = ∠DCE = 60°

∠ODC = ∠BDC = 45°

In △COD,

⇒ ∠COD + ∠ODC + ∠OCD = 180°

⇒ x° + 45° + 60° = 180°

⇒ x° + 105° = 180°

⇒ x° = 180° - 105°

⇒ x° = 75°

Hence, x = 75°.