Question

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

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:

Solution:

Let AB and CD be two equal cords (i.e. AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.

It is now to be proven that the line segments AE = DE and CE = BE

Construction Steps:

Step 1: From the center of the circle, draw a perpendicular to AB i.e. OM ⊥ AB

Step 2: Similarly, draw ON ⊥ CD.

Step 3: Join OE.

Now, the diagram is as follows-

Proof:

From the diagram, it is seen that OM bisects AB and so, OM ⊥ AB

Similarly, ON bisects CD and so, ON ⊥ CD

It is known that AB = CD. So,

AM = ND — (i)

and MB = CN — (ii)

Now, triangles ΔOME and ΔONE are similar by RHS congruency since

OME = ONE (They are perpendiculars)

OE = OE (It is the common side)

OM = ON (AB and CD are equal and so, they are equidistant from the centre)

∴ ΔOME ΔONE

ME = EN (by CPCT) — (iii)

Now, from equations (i) and (ii) we get,

AM+ME = ND+EN

So, AE = ED

Now from equations (ii) and (iii) we get,

MB-ME = CN-EN

So, EB = CE (Hence proved).