Question

In a quadrilateral ABCD, the bisector of $$\angle C$$ and $$\angle D$$ intersect at O. Prove that $$\angle COD=\dfrac{1}{2} \left( \angle A+\angle B\right) $$

Answer 1

Hint: In this question it is given that in a quadrilateral ABCD, the bisector of $$\angle C$$ and $$\angle D$$ intersect at O. So to understand it we have to draw the diagram,

So here we have to prove that, $$\angle COD=\dfrac{1}{2} \left( \angle A+\angle B\right) $$.
So for this we have to use the triangular formula in $\triangle COD$, which is “The sum of all angles of a triangle is $$180^{\circ}$$”. From where, by solving we will be able to get our required solution.

Complete step-by-step answer:
Since, $$\angle OCD$$ and $$\angle ODC$$ are the bisector of $$\angle C$$ and $$\angle D$$, therefore we can write,
$$\angle C=\dfrac{\angle OCD}{2}$$ and $$\angle D=\dfrac{\angle ODC}{2}$$.......(1)
Now since the sum of all angles of a triangle is $180^{\circ}$ Therefore, in $$\triangle COD$$ we can write,
$$\angle COD+\angle OCD+\angle ODC=180^{\circ}$$
$$\Rightarrow \angle COD=180^{\circ}-\angle OCD-\angle ODC$$
$$\Rightarrow \angle COD=180^{\circ}-\left( \angle OCD+\angle ODC\right) $$
$$\Rightarrow \angle COD=180^{\circ}-\left( \dfrac{\angle C}{2} +\dfrac{\angle D}{2} \right) $$ [by using (1)]
$$\Rightarrow \angle COD=180^{\circ}-\dfrac{1}{2} \left( \angle C+\angle D\right) $$.......(2)
Also as we know that for an quadrilateral ABCD, the sum of all angles is $$360^{\circ}$$,
Therefore, we can write,
$$\angle A+\angle B+\angle C+\angle D=360^{\circ}$$
$$\Rightarrow \angle C+\angle D=360^{\circ}-\angle A-\angle B$$
Now putting the above value in equation (2), we get,
$$\angle COD=180^{\circ}-\dfrac{1}{2} \left( 360^{\circ}-\angle A-\angle B\right) $$
$$\Rightarrow \angle COD=180^{\circ}-\dfrac{360^{\circ}}{2} +\dfrac{\angle A}{2} +\dfrac{\angle B}{2}$$
$$\Rightarrow \angle COD=180^{\circ}-180^{\circ}+\dfrac{\angle A}{2} +\dfrac{\angle B}{2}$$
$$\Rightarrow \angle COD=\dfrac{\angle A}{2} +\dfrac{\angle B}{2}$$
$$\Rightarrow \angle COD=\dfrac{1}{2} \left( \angle A+\angle B\right) $$
Hence proved.


Note: While solving this type of problem you need to know that, when a straight line bisects an angle then it means that this line divides the angle in equal two pants. So in the above figure OC is the bisector of the $$\angle C$$, so the generated two angles are $$\angle OCB$$ and $$\angle OCD$$ which are equal, so we can write, $$\angle OCB =\angle OCD$$, also each angles are the half of $$\angle C$$ i.e, $$\angle OCB=\angle OCD=\dfrac{1}{2} \angle C$$.
So you need to use the appropriate information according to the required solution.
Also while writing an angle for example $$\angle OCB$$, you can also write it as $$\angle BCO$$ but the main thi is that, while writing any angle the vertex of the angle should be in the middle.