If two intersecting chords of the circle make an equal angle with the diameter passing through their point of contact to the diameter, prove that the chords are equal.

Hint: The chords can be proved if equal if their distance from the center is equal, which can be done by using congruency.

Complete step by step answer:
The figure according to the condition given in the question is shown below as,

Given: Let’s suppose AB and CD are the two chords of a circle intersecting at Z. PQ is the diameter which passes through their point of intersection as well. The chords are equally inclined to the diameter. It implies that$\angle OLE = \angle OME$.
To prove: the length of chord AB and CD are equal.
Proof:
Construction: Drop two perpendiculars OL and OM from center O to the chord AB and CD respectively, such that they intersect the chords at L and M respectively.
In triangle OLE and triangle OME
$\angle OLE = \angle OME$ (Both are equal to ${90^o}$, as per our construction)
$\angle LEO = \angle MEO$ (Chords are equally inclined with the diameter)
$OE = OE$ (Both are common in the two triangles.)
Hence, $\Delta OLE \cong \Delta OME$ by AAS(Angle-Angle-Side) criteria
Thus, it can be concluded that
$OL = OM$ (By CPCT- Corresponding Parts of Congruent Triangle)
Since the chords are equidistant from the center of the circle, therefore they are equal to each other.
Thus, length of chord AB = length of chord CD.
(Proved)

Note:
The two triangles are said to be congruent if their corresponding sides and corresponding angles are equal.
There are 4 ways by which it can be proved
SSS (Side-Side-Side) Criteria
SAS (Side-Angle-Side)
AAS (Angle-Angle-Side)
ASA (Angle-Side-Angle)
AAA is not a criterion to prove the congruency of the triangle.