Question

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.

Answer 1

Step 1: Given information in the question:

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

To prove:

(i) $MB=MC$and

(ii) $AM=MD$

Step 2: Explanation for the given statement

$AB$ is a chord and $OE\perp AB$ from the center$O$,

We know the perpendicular from the center to a chord bisects the chord.

We obtain,

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

Similarly,

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

According to the question,

$$\begin{gathered} \Rightarrow AB=CD \\ \Rightarrow \frac{1}{2}AB=\frac{1}{2}CD \end{gathered}$$

So, $AE=FD\dots \left(1\right)$

We know, equal chords are equidistance from the center.

And $AB=CD$

So, $OE=OF$

Step 3: Proof of the given conditon

As proved, in right triangles $MOE$ and $MOF$,

hyp. $OE$ = hyp. $OF$ [Common side]

$$\begin{gathered} \Rightarrow OM=OM \\ \Rightarrow \Delta MOE\cong \Delta MOF\mathbf{(}\mathbf{by}\mathbf{}\mathbf{RHS}\mathbf{)} \\ \Rightarrow ME=MF\dots \left(2\right) \end{gathered}$$

Subtracting equations $\left(2\right)$ from $\left(1\right)$, we get

$AE–ME=FD–MF$

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

Again,$AB=CD$ [Given]

And $AM=MD$ [Proved]

$AB–AM=CD–MD$ [Equals subtracted from equal]

So we get, $MB=MC$[Proved part (i)]

Hence it is proved that If two equal chords of a circle intersect, then the parts of one chord are separately equal to the parts of the other chord.