# 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.

**Solution:**

Let AB and CD be the 2 equal chords. AB = CD. Let the chords intersect at point E. Join OE.

To prove AE = CE and BE = DE.

Draw perpendiculars from the center O to the chords. This Perpendicular bisects the chord AB at M and CD at N.

Thus, AM = MB = CN = DN ……(1)

In ∆OME and ∆ONE

∠M = ∠N = 90°

OE = OE

OM = ON (Equal chords are equidistant from the center.)

By RHS criteria, ∆OME and ∆ONE are congruent.

So by CPCT, ME = NE ….. (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.

**Video Solution:**

## 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.

### Maths NCERT Solutions Class 9 - Chapter 10 Exercise 10.4 Question 2:

**Summary:**

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