Question

$\text{In the given parallelogram ABCD, O is}\text{mid-point of AB. CO and DO bisect}\angle BCD\text{and}\angle ADC\text{respectively. Find}\text{the measure of}\angle DOC.$

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

$\text{In parallelogram ABCD,}$

$\angle ADC+\angle BCD={180}^{\circ }$
$\text{(Adjacant angles of a}\text{parallelogram is supplementary)}$

$\Rightarrow \frac{1}{2}\angle ADC+\frac{1}{2}\angle BCD={90}^{\circ }$
$\Rightarrow \angle ODC+\angle DCO={90}^{\circ }...\left(i\right)$

$\text{Also, the sum of all the angles of a}\text{triangle is}{180}^{\circ }.$

$So,\angle DOC+\angle ODC+\angle DCO={180}^{\circ }$
$\text{From}\left(i\right)$
$\Rightarrow \angle DOC+{90}^{\circ }={180}^{\circ }$
$\Rightarrow \angle DOC={180}^{\circ }-{90}^{\circ }$
$\Rightarrow \angle DOC={90}^{\circ }$