Question

If 2 equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.

Answer 1

Hint:-Let us name the two chords of the circle as AB and CD which are intersecting at P and let the centre of the circle be O.

Complete step-by-step answer:

We know that: AB and CD are chords of a circle with centre O. AB and CD intersect at P. Also, AB = CD.

To prove: AP = PD and PB = CP.

Construction: Draw OM perpendicular to AB and ON perpendicular CD. Join OP.

AM = MB = ½ AB (Perpendicular bisecting the chord)

CN = ND = ½ CD (Perpendicular bisecting the chord)

AM = ND and MB = CN (As AB = CD)

In triangle OMP and ONP, we have,

OM = MN (Equal chords are equidistant from the centre)

Angle OMP = Angle ONP (90⁰)

OP is common. Thus triangle OMP and ONP are congruent (RHS).

MP = PN (By CPCT)

Therefore, AM + MP = ND + PN

or, AP = PD (Equation (i))

As MB = CN and MP = PN,

MB - MP = CN - PN

Therefore, PB = CP (Equation (ii))

Hence, proven.

Note:-CPCT stands for ‘Corresponding parts of Congruent triangles’. CPCT theorem states that if two or more triangles which are congruent to each other are taken then the corresponding angles and the sides of the triangles are also congruent to each other.