Question

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Answer 1

Given: Consider AB and CD are two equal chords of a circle, which meet at point E.

To prove: AE = CE and BE = DE

Construction: Draw OM ⊥ AB and ON ⊥ CD and join OE where O is the centre of circle.

Proof: In ΔOME and ΔONE,

OM = ON  ...[Equal chords are equidistant from the centre]

OE = OE   ...[Common side]

And ∠OME = ∠ONE   ...[Each 90°]

 ∴ ΔOME ≅ ΔONE   ...[By RHS congruence rule]

⇒ EM = EN   [By CPCT]  ...(i)

Now, AB = CD

On dividing both sides by 2, we get

`(AB)/2 = (CD)/2` ⇒ AM = CN   ...(ii) [Since, perpendicular drawn from centre of circle to chord bisects the chord i.e., AM = MB and CN = ND]

On adding equations (i) and (ii), we get

EM + AM = EN + CN 

⇒ AE = CE   ...(iii)

Now, AB = CD

On subtracting both sides by AE, we get

AB – AE = CD – AE

⇒ BE = CD – CE  ...[From equation (iii)]

⇒ BE = DE 

Hence proved.