Question

Prove that the chords equidistant from the center of a circle are equal in length.

Answer 1

Hint: First of all, draw a circle with center O and chords AB and CD at a distance of OX and OY from the center respectively. Now, join O to A and O to D. Now prove $$\mathrm{\Delta }AOX$$ and $$\mathrm{\Delta }DOY$$ congruent, and from this prove AX = DY. Now, from this prove AB = CD by using the theorem that perpendiculars from the center bisect the chord.

Complete step-by-step answer:

In this question, we have to prove that the chords equidistant from the center of a circle are equal in length. First of all, construct a circle with center O

In the above circle, AB and CD are two chords of the circle where OX is the distance from the center of the chord AB that is $$OX\mathrm{\perp }AB$$. Also, OY is the distance from the center of the chord CD that is $$OY\mathrm{\perp }CD$$. We are given that the two chords are equidistant from the center of the circle. So, OX = OY. Now, we have to prove that the lengths of the given chords are equal or AB = CD. Let us join point A to O and point D to O.

Consider $$\mathrm{\Delta }AOX$$ and $$\mathrm{\Delta }DOY$$. We get, both $$\mathrm{\angle }OXA$$ and $$\mathrm{\angle }OYD$$ are equal to $${90}^o$$. So,

$$\mathrm{\angle }OXA=\mathrm{\angle }OYD.....\left(i\right)$$

We know that in a circle, all radii are equal. So,

OA = OD …..(ii)

Also, we are given that OX = OY……(iii)

So, from right angle hypotenuse side (RHS) congruence criteria, we get,

$$\mathrm{\Delta }AOX\cong \mathrm{\Delta }DOY$$

We know that the corresponding parts of the congruent triangles are equal. So, we get,

AX = DY……(iv)

We know that the perpendicular from the center to a chord bisects the chord. So, we get,

For chord AB, X bisects AB, so AB = 2 AX.
$$\Rightarrow AX={\displaystyle \frac{AB}{2}}.....\left(v\right)$$

For chord CD, Y bisects CD, so, CD = 2 DY

$$\Rightarrow DY={\displaystyle \frac{CD}{2}}.....\left(vi\right)$$

Now by substituting the values of AX and DY from equation (v) and (vi) in equation (iv), we get,

$${\displaystyle \frac{AB}{2}}={\displaystyle \frac{CD}{2}}$$

AB = CD [Hence proved]

So, we have proved that chords equidistant from the center of the circle is equal in length.

Note: In this question, students must remember all the theorems discussed in the above solution as they are very useful while solving the questions related to the circle. Also, students can check these theorems practically by constructing a circle of a certain radius and making equidistant chords and perpendiculars on it from the center. Now, measure the lengths of both the chords if they are equal or not.