In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠MAD = x and ∠BAC = y :

(i) express ∠AMD in terms of x.

(ii) express ∠ABD in terms of y.

(iii) prove that : x = y.

Mark the point of intersection of AB and CD as L.

(i) In △AMD,

MA = MD [Radius of circle]

∴ ∠MAD = ∠MDA = x.

In △AMD,

⇒ ∠MAD + ∠MDA + ∠AMD = 180° [Angle sum property of triangle]

⇒ x + x + ∠AMD = 180°

⇒ ∠AMD = 180° - 2x.

Hence, ∠AMD = 180° - 2x.

(ii) Let the perpendicular chords AB and CD intersect each other at L.

From figure,

∠ALC = 90°.

In △ALC,

⇒ ∠LAC + ∠LCA + ∠ALC = 180° [Angle sum property of triangle]

⇒ y + ∠DCA + 90° = 180°

⇒ ∠DCA = 180° - 90° - y

⇒ ∠DCA = 90° - y

From figure,

∠ABD = ∠DCA [Angles in same segment are equal]

∴ ∠ABD = 90° - y.

Hence, ∠ABD = 90° - y.

(iii) We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AMD = 2∠ABD

∠ABD = $\dfrac{1}{2}$ ∠AMD = $\dfrac{1}{2} \times (180° - 2x)$ = 90° - x.

We have,

∠ABD = 90° - y and ∠ABD = 90° - x

⇒ 90° - y = 90° - x

⇒ x = y.

Hence, proved that x = y.