Question

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

Answer 1

In parallelogram ABCD,

$\angle ADC+\angle BCD=180{}^{\circ }$
Adjacent angles of a parallelogram is supplementary

$\frac{\angle ADC}{2}+\frac{\angle BCD}{2}=90{}^{\circ }$

$\angle ODC+\angle DCO=90{}^{\circ }....\left(i\right)$
Also, the sum of all the angles of a triangle is }180°.

So, $\angle DOC+\angle ODC+\angle DCO=180{}^{\circ }$
From (i)

$\angle DOC+90{}^{\circ }=180{}^{\circ }$

$\angle DOC=180{}^{\circ }-90{}^{\circ }$

$\angle DOC=90{}^{\circ }$