Question

In the given figure, ΔABC is isosceles triangle with $AB=ACand\angle ABC={40}^{\circ }$. The difference of $\angle BDCand\angle BEC$ is equal to

• A.
  ${35}^{\circ }$
• B.
  ${20}^{\circ }$
• C.
  ${30}^{\circ }$
• D.
  ${25}^{\circ }$

Answer 1

The correct option is B

${20}^{\circ }$
Given that AB = AC.

Since angles opposite to equal sides are equal.



$\therefore \angle ABC=\angle ACB\Rightarrow \angle ACB={40}^{\circ }(\because \angle ABC={40}^{\circ })Now,in\Delta ABC,\text{by Angle sum property,}\angle BAC+\angle ABC+\angle ACB={180}^{\circ }\Rightarrow \angle BAC+{40}^{\circ }+{40}^{\circ }={180}^{\circ }\Rightarrow \angle BAC={180}^{\circ }–{80}^{\circ }={100}^{\circ }$

Now, angles subtended by a chord in the same segment are equal.

$\Rightarrow \angle BDC=\angle BAC=100º$

Since in a cyclic quadrilateral, opposite angles are supplementary.

$\therefore \angle BEC+\angle BDC={180}^{\circ }\Rightarrow \angle BEC={180}^{\circ }–{100}^{\circ }={80}^{\circ }\Rightarrow \angle BDC–\angle BEC={100}^{\circ }–{80}^{\circ }={20}^{\circ }$

Hence, the correct answer is option (b).