Sum of angles of a quadrilateral is 360°
In the quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
∠A + ∠B = 360° - (∠C + ∠D)
\(\frac{1}{2}\)∠A + ∠B = \(\frac{1}{2}\){360° - (∠C + ∠D)}
\(\frac{1}{2}\)∠A + ∠B = 180° - \(\frac{1}{2}\)(∠C + ∠D) …….(i)
Now in Δ DOC
\(\frac{1}{2}\)∠D + \(\frac{1}{2}\)∠C + ∠COD = 180° [Sum of angles of a triangle is 180°]
\(\frac{1}{2}\)(∠D + ∠C) + ∠COD = 180°
∠COD = 180° - \(\frac{1}{2}\)(∠C + ∠D) ……….(ii)
From above equations (i) and (ii) RHS is equal therefore LHS will also be equal.
Therefore ∠COD = \(\frac{1}{2}\)(∠A + ∠B)
