In the adjoining quadrilateral $A B C D, A O$ and $B O$ are the bisectors of $∠ A and ∠ B$ respectively. Prove that $∠ AOB = \frac{1}{2} (∠ C + ∠ D) .$

Open in App

Solution

We know,
$∠ A + ∠ B + ∠ C + ∠ D = 360$ (sum of angles of quadrilateral)

$∠ A + ∠ B = 360 - (∠ C + ∠ D) - - - (1)$

In $△ A O B$ , According to $Q u e s t i o n$
$\frac{1}{2} (∠ A + ∠ B) + ∠ O = 180$ (sum of angles of triangle)

$∠ O = 180 - \frac{1}{2} (∠ A + ∠ B) - - - (2)$

Substitute value of $(∠ A + ∠ B)$ in $(2)$ from $(1)$

$∠ O = 180 - \frac{1}{2} [360 - (∠ C + ∠ D)]$

$∠ A O B = ∠ O = \frac{1}{2} (∠ C + ∠ D)$
Hence proved

Suggest Corrections

Similar questions

Q. In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that

∠

AOB

= \frac{1}{2} {∠ C + ∠ D}

In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively.

Show that :

$∠ A O B = \frac{1}{2} (∠ C + ∠ D)$

In a quadrilateral $A B C D, A O$ and $B O$ are the bisectors of $∠ A a n d ∠ B$ respectively, $∠ C = 70^{\circ} a n d ∠ D = 30^{\circ}$ . Then, $∠ A O B =$ ?

(a) $40^{\circ}$

(b) $50^{\circ}$

(d) $100^{\circ}$

Q. In a quadrilateral ABCD AO and bo are the bisector of angle A and Angle B respectively prove that angle aob is equals to half of angle C plus half of angle d

A B C D

is a quadrilateral.

A O

and

B O

are the angle bisectors of angles

A

and

B

which meet at

O

. If

∠ C = 70^{o}, ∠ D = 50^{o}

, find

∠ A O B

Join BYJU'S Learning Program

In the adjoining quadrilateral ABCD,AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 12(∠C + ∠D).

In the adjoining quadrilateral $A B C D, A O$ and $B O$ are the bisectors of $∠ A and ∠ B$ respectively. Prove that $∠ AOB = \frac{1}{2} (∠ C + ∠ D) .$