Question

If two equal chords of a circle intersect, prove that the parts of one chords are separately equal to the parts of the other chord.

Answer 1

$AB$ and $CD$ are two equal chords of a circle with centre $O$, intersect each other at $M$.

We have to prove that,

(i) $MB=MC$ abd

(ii) $AM=MD$

Refer image,

$AB$ is a chord and $OE\perp$ to it from the centre $O$,

$\therefore AE=\frac{1}{2}AB$

[$\because$ Perpendicular from the centre to a chord bisects the chord]

Similarly $\therefore FD=\frac{1}{2}CD$

As $AB=CD\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$ [GIven]

So, $AE=FD$.........(1)

Since equal chords are equidistant from the centre.

So, $OE=OF[\because AB=CD]$

Now, in right $\Delta MOE$ and $MOF$

hyp $OE=hyp.OF$ [Proved above]

$OM=OM$ [Common side]

$\therefore \Delta MOE\cong \Delta MOF$ [By RHS congruence]

$\therefore ME=MF$............(2)

Subtracting (2) from (1), we get

$AE-ME=FD-MF$

$\Rightarrow AM=MD$ [Proved part (ii)]

Again, $AB=CD$ [Given]

and 4$AM=MD$ [Proved]

$\therefore AB-AM=CD-MD$ [Equals subtracted from equal]

Hence, $MB=MC$ [Proved part (i)]