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

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)

in progress 0
Lyla 3 weeks 2021-08-21T23:41:43+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-08-21T23:43:23+00:00

    Answer:

    Brainly.in

    What is your question?

    1

    karansandhu155

    08.12.2014

    Math

    Secondary School

    +5 pts

    Answered

    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)

    1

    SEE ANSWER

    Log in to add comment

    Answers

    Me · Beginner

    Know the answer? Add it here!

    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.

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )