F.5 MATHS

Let ∠OEC = a and ∠OCE = b

By the property of tangent of circle, EO bisects ∠AEC.
∠OEA = ∠OEC = a
Hence, ∠AEC = ∠OEA + ∠OEC = 2a

Similarly, CO bisects ∠BCE.
∠OCB = ∠OCE = b
Hence, ∠BCE = ∠OCB + ∠OCE = 2b

∠AEC + ∠BCE = 180° (int. ∠s, AE // BC)
2a + 2b = 180°
a + b = 90°

In ΔOEC:
∠EOC + ∠OEC + ∠OCE = 180° (∠ sum of Δ)
∠EOC + a + b = 180°
∠EOC + 90° = 180°
∠EOC = 90°