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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 :
- express ∠AMD in terms of x.
- express ∠ABD in terms of y.
- 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∠AMD at 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^circ - 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
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