In the figure, AO and BO are the bisectors of two adjacent angles A an

1.	In the figure, AO and BO are the bisectors of two adjacent angles A and B of a quad. ABCD. Prove that 2∠AOB = ∠C + ∠D
Answer» Given: In quad. ABCD the bisectors of ∠A and ∠B intersect at O. To prove: 2∠AOB = ∠C + ∠D Proof: In quad. ABCD ∠A + ∠B + ∠C + ∠D = 360° …(i) (angle sum property of a quad.) Also in ΔAOB ∠OAB + ∠OBA + ∠AOB = 180° (by angle sum property of a Δ) ⇒ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B + ∠AOB = 180° ⇒ ∠A + ∠B + 2∠AOB = 360°…(ii) From (i) and (ii), we get ∠A + ∠B + ∠C + ∠D = ∠A + ∠B + 2∠AOB ⇒ 2∠AOB = ∠C + ∠D

In the figure, AO and BO are the bisectors of two adjacent angles A and B of a quad. ABCD. Prove that 2∠AOB = ∠C + ∠D

Answer»

Given: In quad. ABCD the bisectors of ∠A and ∠B intersect at O.

To prove: 2∠AOB = ∠C + ∠D

Proof: In quad. ABCD

∠A + ∠B + ∠C + ∠D = 360° …(i) (angle sum property of a quad.)

Also in ΔAOB

∠OAB + ∠OBA + ∠AOB = 180° (by angle sum property of a Δ)

⇒ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B + ∠AOB = 180°

⇒ ∠A + ∠B + 2∠AOB = 360°…(ii)

From (i) and (ii), we get

∠A + ∠B + ∠C + ∠D = ∠A + ∠B + 2∠AOB

⇒ 2∠AOB = ∠C + ∠D