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  • 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.

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.

Answers (1)

Given: Two equal chords of a circle intersect

Let us construct a circle with centre O.

Its two equal chords are AB and CD which intersect at E as shown in the figure.

To prove: AE = DE and CE = BE

Proof:

Draw OM\perp AB and ON \perpCD

Join OE

In \triangleOME and \triangleONE,

OM = ON (Equals chords of a circle are equidistant from the centre.)

OE = OE (Common)

\angleOME =\angleONE (90o)

\because \triangleOME \cong \triangleONE (RHS congruence)

ME = NE  (C.P.C.T.) ..…(i)

Now as AB = CD

\frac{1}{2}AB =\frac{1}{2}CD

AM = DN..…(ii)

Adding (i) and (ii)

AM + ME = DN + NE           

Þ AE = DE

Also, AB – AE = CD – DE (\becauseAB = CD)

 BE = CE

Hence proved

Posted by

infoexpert24

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