Question

In the given figure, O is the centre of the circle. Chord AB is parallel to chord CD and CB is a diameter. Prove that arc AC = arc BD. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Given: $ChordAB\parallel ChordCD$ and COB is a diameter.

To prove: Arc AC = Arc BD.

Proof:

$\angle AOC=2\angle ABC$ [Angle subtended by the chord at the centre is double the angle at any point on the circumference.]

$⟹\angle ABC=\frac{1}{2}\angle AOC$ ...(i)

$\angle BOD=2\angle BCD$ [Angle subtended by the chord at the centre is double the angle at any point on the circumference.]

$⟹\angle BCD=\frac{1}{2}\angle BOD$ ...(ii)

$\angle ABC=\angle BCD$ [Alternate $\angle$s, as $AB\parallel CD$]

$⟹\frac{1}{2}\angle AOC=\frac{1}{2}\angle BOD$ [From (i) and (ii)]

$⟹\angle AOC=\angle BOD$

$⟹$ Arc AC = arc BD in a circle, the arcs subtending equal $\angle$s at the centre are equal

Hence, Proved.