# in a quadrilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively. Prove that angle AOB = 1/2 (angle C+ angle D)

Given, AO and BO are the bisectors of angle A and angle B respectively.

∴ ∠1 = ∠4 and ∠3 = ∠5 ... (1)

To prove: ∠2 = (∠C + ∠D) Proof:

∠A + ∠B + ∠C + ∠D = 360° (∠A  + ∠B + ∠C + ∠D) = 180°  ... (2)

Now in ΔAOB

∠1 + ∠2 + ∠3 = 180° ... (3)

equating (2) and (3), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 + (∠C + ∠D)

∴ ∠2 = [∠C + ∠D]

Hence proved

• 46

thnx..i got another way for finding the ans..but this one is easy

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Given, AO and BO are the bisectors of angle A and angle B respectively.

∴ ∠1 = ∠4 and ∠3 = ∠5 ... (1)

To prove: ∠2 = (∠C + ∠D) Proof:

∠A + ∠B + ∠C + ∠D = 360� (∠A + ∠B + ∠C + ∠D) = 180� ... (2)

Now in ΔAOB

∠1 + ∠2 + ∠3 = 180� ... (3)

equating (2) and (3), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 + (∠C + ∠D)

∴ ∠2 = [∠C + ∠D]

Hence proved

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1) In any quadrilateral, sum of its four angles = 360°

As such here in the quadrilateral ABCD also, <A + <B + <C + <D = 360°

==> <C + <D = 360° - (<A + <B)

2) Dividing the above by 2,

(1/2)(<C + <D) = 180° - (1/2)(<A + <B) ------- (i)

3) In the triangle AOB, <AOB = 180° - (1/2)*(<A + <B) ------- (ii) [Since given AO & Bo are bisectors of angles A & B respectively]

4) Thus from (i) & (ii) above,

<AOB = (1/2)(<C + <D) [Proved]
• 4 • 29
just add up angle D and angle C and divide by 2 to get your final answer.
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