Prove that if the two arms of an angle are perpendicular to the two arms of another angle.

Consider be angles AOB and ACB

Given 0A perpendicular to A0, also 0B perpendicular to BO

To Prove: ∠AOB + ∠ACB = 180° (or) ∠AOB + ∠ACB = 180°

Proof: In a quadrilateral = ∠A + ∠O + ∠B + ∠C = 360°

[Sum of angles of quadrilateral is 360]

⟹ 180 + O + B + C = 360

⟹ O + C = 360 - 180

Hence AOB + ACB = 180 .... (1)

Also, B + ACB = 180

⟹ ACB = 180 - 90 = ACES = 90°   .... (2)

From (i) and (ii), ACB = A0B = 90

Hence, the angles are equal as well as supplementary.