In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If ∠OAB = 25°, calculate ∠EBC.

OA = OB (Radii of a circle)
Thus, ∠OBA = ∠OAB = 25°
Join OB.

Now in ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)
$\Rightarrow$25° + 25° + ∠AOB = 180°
$\Rightarrow$50° + ∠AOB = 180°
$\Rightarrow$∠AOB = (180° – 50°) = 130°

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., ∠AOB = 2∠ACB
$\Rightarrow$$\angle ACB\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\angle }AOB=\left(\frac{1}{2}\times 130°\right)=65°$
Here,∠ACB = ∠ECB
∴ ∠ECB = 65° ...(i)

Considering the right angled ΔBEC, we have:
∠EBC + ∠BEC + ∠ECB = 180° (Angle sum property of a triangle)
$\Rightarrow$∠EBC + 90° + 65° = 180° [From(i)]
$\Rightarrow$∠EBC = (180° – 155°) = 25°
Hence, ∠EBC = 25°