Question

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.

Answer 1

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.