#### Question

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.

#### Solution

In the figure, M is the centre of the circle.

Chords AB and CD are perpendicular to each other at L.

∠MAD = x and ∠BAC = y

(i) In ∆AMD,

MA = MD

∴ ∠MAD = ∠MDA = x

But in ∆AMD,

∠MAD + ∠MDA + ∠AMD = 180°

⇒ x + x + ∠AMD = 180°

⇒ 2x + ∠AMD = 180°

⇒ ∠AMD =180° - 2x

(ii) ∴ Arc AD ∠AMDat the centre and ∠ABD at the remaining

(Angle in the same segment)

(Angle at the centre is double the angle at the circumference subtended by the same chord)

⇒ ∠AMD = 2∠ABD

⇒`∠ABD = 1/2 (180° - 2x )`

⇒∠ABD = 90° - x

AB ⊥ CD, ∠ALC = 90°

In ∆ALC,

∴ ∠LAC +∠LCA = 90°

⇒ ∠BAC + ∠DAC = 90°

⇒ y + ∠DAC = 90°

∴ ∠DAC = 90° - y

We have, ∠DAC = ∠ABD [Angles in the same segment]

∴ ∠ABD = 90° - y

(iii) we have, ∠ABD =90° - y and ∠ABD = 90° - x [proved]

∴ 90° - x 90° - y

⇒ x = y