# Ex.10.4 Q2 Circles Solution - NCERT Maths Class 9

## Question

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

## Text Solution

**What is known?**

Chords are equal. Chords intersect at a point within the circle.

**What is unknown?**

Proof of corresponding segments of equal chords are equal.

**Reasoning:**

Equal chords are equidistant from the centre. Using this and Right angled-Hypotenuse-Side (**RHS**) criteria and Corresponding parts of congruent triangles (**CPCT**) we prove the statement.

**Steps:**

Let \({AB}\) and \({CD}\) be the \(2\) equal chords. \({AB = CD.}\)

Let the chords intersect at point \({E}\). Join \({OE.}\)

To prove \(\begin{align} {AE = CE \, {\rm{and}} \, BE = DE.} \end{align}\)

Draw perpendiculars from the center to the chords. Perpendicular bisects the chord \({AB}\) at \({M}\) and \({CD}\) at \({N.}\)

\(\begin{align} {AM = MB = CN = DN ……(1)} \end{align}\)

\(\begin{align} {\rm{In}} \;\Delta {OME}\; {\rm{and}} \;\Delta {ONE}\end{align}\)

\[\begin{align}\angle {M}&=\angle {N}=90^{\circ}\\ {OE}&={OE}\\{OM}&={ON} \quad\\\end{align}\]

Equal chords are equidistant from the centre.

By **RHS** criteria, \(\Delta {OME}\) and \(\Delta {ONE}\) are congruent.

So by \(\text{CPCT}, {ME}={NE} \ldots \ldots(2)\)

We know that: \({CE}={CN}+{NE}\) and

\({AE}={AM}+{ME}\)

From (**\(1\)**) and (**\(2\)**), it is evident \({CE = AE}\)

\({DE}={CD}-{CE}\) and

\({BE}={AB}-{AE}\)

\({AB}\) and \({CD}\) are equal, \({CE}\) and \({AE}\) are equal. So \({DE}\) and \({BE}\) are also equal.

It is proved corresponding segments of equal chords are equal.