Question

$\huge \bf \: Question$
In a Quadrilateral ABCD, CO and DO are bisector of $\sf \angle C$ and $\sf \angle D$
Proof :-
$\bf \angle \: COD \: = \dfrac{1}{2}( \angle \: A + \angle \: B )$
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Answer 1

In Triangle COD

➤ By the Angle Sum Property (Angle sum property of triangle states that Interior angles of a Triangles sums up to 180°),

$\to\sf{\angle 1\:+\:\angle 2\:+\:\angle COD\:=\:180^\circ}$

$\to\sf{\angle 1\:+\:\angle COD\:=\:180^\circ\:-\:\angle 2}$

$\to\sf{\angle COD\:=\:180^\circ\:-\:\angle 1\:-\:\angle 2}$

✯ Which can be written as

$\to\sf{\angle COD\:=\:180^\circ\:-\:(\angle 1\:+\:\angle 2)}$

$\to\sf{\angle COD\:=\:180^\circ\:-\:\left(\dfrac{1}{2}\angle C\:+\:\dfrac{1}{2}\angle D\right)}$

✯ Taking 1/2 common

$\to\sf{\angle COD=180^{\circ}-\dfrac{1}{2}\left(\angle C+\angle D\right)\:\:\:\bf{.......1}}$

➤ We know A quadrilateral also has four interior angles and their sum equals 360° which is also known as Angle sum property of a quadrilateral

$\to\sf{\angle A+\angle B+\angle C+\angle D=360^{\circ}}$

$\to\sf{\angle B+\angle C+\angle D=360^{\circ}-\angle A}$

$\to\sf{\angle C+\angle D=360^{\circ}-\angle A-\angle B}$

✯ Which can be written as

$\to\sf{\angle C+\angle D=360^{\circ}-(\angle A+\angle B)}$

➤ Here We Got the value of $\sf{\angle C\:+\:\angle D}$ and now keep/substitute this value in Equation 1

$\to\sf{\angle COD=180^{\circ}-\dfrac{1}{2}\left(\angle C+\angle D\right)}$

$\to\sf{\angle COD=180^{\circ}-\dfrac{1}{2}[360^{\circ}-(\angle A+\angle B)]}$

$\to\sf{\angle COD=180^{\circ \:}-\dfrac{1}{2}\left(360^{\circ \:}-\angle A-\angle B\right)}$

$\to\sf{\displaystyle \angle COD=180^\circ-\left(\left(-\frac{1}{2}\right)\times \:360^{\circ \:}+\left(-\frac{1}{2}\right)\left(-\angle A\right)+\left(-\frac{1}{2}\right)\left(-\angle B\right)\right)}$

$\to\sf{\displaystyle \angle COD=180^\circ-\left(360^{\circ \:}\times\frac{1}{2}+\frac{1}{2}\angle A+\frac{1}{2}\angle B\right)}$

$\to\sf{\displaystyle \angle COD=180^{\circ \:}-180^{\circ \:}+\frac{\angle A}{2}+\frac{\angle B}{2}}$

$\boxed{\boxed{\to\sf{\angle C O D=\dfrac{1}{2}(\angle A+\angle B)}}}$

Hence Proved !!

Answer 2

Answer:

angle COD=1/2(angle A+angle B)

Step-by-step explanation:

$angle1 + angle2 = 180$

=angle 2 +angleCOD = 180 -angle 2

=angleCOD = 180 -angle2-angle 1

=180 ( angle 1 -angle 2)

= 180 ( 1/2 angle A+1/2angleB)

=angleCOD=1/2(angle A +angle B)