Question

Please give me answer of this question​

Answer 1

Answer:

Step-by-step explanation:

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 !!