Let the two equal chords be $AB$ and $CD$, intersecting at point $E$. Let $O$ be the centre of the circle. We draw perpendiculars $OM$ and $ON$ from $O$ to $AB$ and $CD$ respectively. We are given that $AB = CD$.

Since a perpendicular from the centre to a chord bisects it, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $CD$. Thus,

$$AM = MB = \frac{1}{2}AB$$

and

$$CN = ND = \frac{1}{2}CD$$

Because $AB = CD$, we get $AM = CN$ and $MB = ND$. Also, chords of equal length lie at equal distance from the centre, so $OM = ON$.

Now, consider the right triangles $\triangle OME$ and $\triangle ONE$. We have $OM = ON$, $OE = OE$ (common side), and $\angle OME = \angle ONE = 90°$. By the RHS congruence rule, $\triangle OME \cong \triangle ONE$. Hence, their corresponding parts are equal, giving $ME = NE$.

Since $M$ and $N$ are midpoints, the segments of chord $AB$ are:

$$AE = AM + ME$$

and

$$EB = MB - ME$$

(assuming $E$ is between $M$ and $B$). Similarly, for chord $CD$:

$$CE = CN + NE$$

and

$$ED = ND - NE$$

Substituting $AM = CN$ and $ME = NE$, we get $AE = CE$. Similarly, substituting $MB = ND$ and $ME = NE$, we get $EB = ED$.

∴ The line segments of one chord are equal to the corresponding line segments of the other chord.