Question

In figure, $ABCD$ is a cyclic quadrilateral whose diagonals intersect each other at $O$. If $\angle ACB={50}^0$ and $\angle ABC={110}^0$, then $\angle BDC$ will be :

• A. ${110}^0$
• B. ${60}^0$
• C. ${40}^0$
• D. ${20}^0$

Answer 1

The correct option is D ${20}^0$

$\because$ Angles formed on same base $AB$,
$\angle ADB=\angle ACB$
$⟹$ $\angle ADB={50}^0$...[because $\angle ACB={50}^0$].

We know that sum of opposite angles of cyclic quadrilateral is ${180}^0$
$\therefore \angle ADC+\angle ABC={180}^0$
$⟹$ $\angle ADC+{110}^0={180}^0$....[because $\angle ABC={110}^0]$
$⟹$ $\angle ADC={180}^0–{110}^0={70}^0$.

Now $\angle ADC=\angle ADB+\angle BDC$
$⟹$ ${70}^0={50}^0+\angle BDC$
$⟹$ ${70}^0–{50}^0=\angle BDC$
$⟹$ ${20}^0=\angle BDC$.

Thus, $\angle BDC={20}^0$.
Therefore, option $D$ is correct.