Question

In the figure BD = DC and $\angle DBC={25}^{\circ }$. What is the value of $\angle BAC$?

• A. ${30}^{\circ }$
• B. ${100}^{\circ }$
• C. ${50}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${50}^{\circ }$

Since, BD = DC in $\Delta BDC$
Therefore, $\Delta BDC$ is an isosceles $\Delta .$
Now, by the property of isosceles $\Delta$.
$\angle DBC=\angle DCB={25}^{\circ }$
Now, $\angle BDC={180}^{\circ }-{50}^{\circ }={130}^{\circ }$ (Sum of angles of $\Delta ={180}^{\circ })$
Now, ABCD is a cyclic quadrilateral
$\Rightarrow \angle D+\angle A={180}^{\circ }$
${130}^{\circ }+\angle A={180}^{\circ }$
$\Rightarrow \angle A=\angle BAC={50}^{\circ }$