If two chords of a circle are equally inclined to the diameter through their point of intersection, then the chords are equal.

Concept: It is known that the chords that are equidistant from the center of a circle have the same length.

Here, $\text{OL}\perp \text{AB}$ and  $\text{OM}\perp \text{AC}$$\text{OL}=\text{OM}$

Then, $\angle \text{OLA}=\angle \text{OMA}$

and $\text{OA}=\text{OA}$

We have  $\angle \text{OAL}=\angle \text{OAM}$

By AAS postulate,

$\Delta \text{OLA}=\Delta \text{OMA}$

By CPCTC,

$\text{OL}=\text{OM}$

Chords are equidistant from center O.

Therefore $\text{AB=CD}$.

Hence, the given statement is true.