Question

In the following figure, if O is the centre of the circle then $\angle DOE$ equals

• A. ${45}^{\circ }$
• B. ${50}^{\circ }$
• C. ${75}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D

${90}^{\circ }$

Given, O is the centre of the circle. So, AB must be a diameter of this circle.

We know that measuer of the angle in a semicircle is 90°.

Therefore $\angle ACB={90}^{\circ }$.

Now, in triangle DCE and triangle DOE,

CD = DO and CE = OE (given)

DE = DE (common side)

Then by SSS postulate, $△DCE\cong △DOE$.

Hence by C.P.C.T.C., $\angle DCE=\angle DOE$

Therefore $\angle DOE=\angle DCE=\angle ACB={90}^{\circ }$