Question

In a quadrilateral ABCD,AO and BO are the bisectors of angle A and angle B.prove that angle AOB =half of (angle C+angle D)

Answer 1

Answer:

Step-by-step explanation: I will tell you 2 ways.

First way :-

In triangleAOB,

A+B+AOB=180 A=1,B=2

AOB=180-(A+B)

A+B+C+D=360

1/2(A+B)+1/2(C+D)=180

1+2+!/2(C+D)=180

1/2(C+D)=180-(1+2)

=AOB

1/2(C+D)=ANG AOB

HENCE PROVED

Second Way: -

∠a + ∠b = 180°

1/2 (∠a + ∠b) = 90°

in tri. AOB, ∠oab + oba +∠aob = 180°

now, ∠aob + 90° = 180° (∵ ∠oab + ∠oba = 90° )

so ∠aob = 180° - 90° = 90°

now, ∠d + ∠c = 180°

so, 1/2 (∠d + ∠c) = 90°

therefore,∠aob = 1/2(∠d + ∠c) = 90°

hence, proved

Thank you Hope you all understand.

Answer 2

Answer:

gay hamari mata hai hame kuch nhi aata hai