Question

ii) Prove : Suppose two chords of a circle intersect each other in the interior of the
circle then the product of the lengths of the two segments of one chord is equal
to the product of the lengths of the two segments of the other chord.​

Answer 1

Answer:

Given AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.

To prove : AP = PD and PB = CP.

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

AM = MB = 1/2AB (Perpendicular bisecting the chord)

CN = ND = 1/2CD (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)

<OMP = <ONP (90⁰)

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

MP = PN (cpct)

So, AM + MP = ND + PN

or, AP = PD (i)

As MB = CN and MP = PN,

MB - MP = CN - PN

= PB = CP (ii)

Hope that helps !!

Answer 2

Answer:

oh my that seems like a little too much!