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# 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) - ICSE Class 10 - Mathematics

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ConceptChord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof)

#### 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

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Solution 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) Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).
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