Question

In a quadrilateral ABCD . AO and OB are bisectors of angle A and angle B . PT angel AOB = 1/2( angle C + angle D)

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karansandhu155

08.12.2014

Math

Secondary School

+5 pts

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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)

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manojdwivedi030

manojdwivedi030 Ambitious

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 understand.