# 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.

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**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 CD and EF are the two chords of a circle intersecting at Z. AB is the diameter which passes through their point of intersection as well. The chords are equally inclined to the diameter. It implies that$\angle XZO = \angle YZO$.

To prove: the length of the chord CD and EF are equal.

Proof:Construction: Drop two perpendiculars from center O to the chord CD and EF respectively, such that they intersect the chords at X and Y respectively.

In triangle OXZ and triangle OYZ

$\angle OXZ = \angle OYZ$ (Both are equal to ${90^o}$, as per our construction)

$\angle XZO = \angle YZO$ (Chords are equally inclined with the diameter)

$OZ = OZ$ (Both are common in the two triangles.)

Hence, $\Delta OXZ \cong \Delta OYZ$ by AAS(Angle-Angle-Side) criteria

Thus, it can be concluded that

$OX = OY$ (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, the length of chord CD = length of chord EF.

(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.