Equal Chords Are Equidistant

We will learn another important property of circles today.

Theorem:

$Chords\ of\ equal\ length\ are\ at\ equal\ distance\ from\ the\ center\ of\ the\ circle.$

Just a quick recap of what we mean by distance of a line from a point. The distance of a line from a point is defined as the length of the perpendicular drawn from that point on the line. Look at the diagram below.

$AB$ and $CD$ are two chords of equal length of the given circle with center $O$. We have drawn perpendiculars from the center $O$ on $AB$ at point $P$ and on $CD$ at point $Q$. We should remember that $OP = OQ$. We will see the formal proof now.

-----------book page break-----------

Given:

A circle with center $O$, has two equal length chords $AB$ and $CD$. Perpendiculars are drawn from the center $O$ to these chords at points $P$ and $Q$ respectively.

Required To Prove:

$OP = OQ$.

Construction:

We draw lines $OA$, $OB$, $OC$ and $OD$.

Proof:

In $\triangle OAB$ and $\triangle OCD$:

$OA = OC$ (radii of the same circle)

$OB = OD$ (radii of the same circle)

$AC = BD$ (given).

$\texttip{\therefore}{therefore} \triangle OAC \texttip{\cong}{congruent to} \triangle OBD$

$\texttip{\therefore}{therefore} \angle PBO = \angle QDO$

-----------book page break-----------

$\triangle OAB$ is an isosceles triangle since $OA = OB$. We know that the perpendicular draw to the unequal side from the opposite vertex bisects the side.

$\texttip{\therefore}{therefore} BP = AP = \dfrac{1}{2} \times AB$

$\triangle OCD$ is an isosceles triangle since $OC = OD$. Because of the same reason above,

$DQ = CQ = \dfrac{1}{2} \times CD$

$AB = CD$ (given)

$\texttip{\therefore}{therefore} \dfrac{1}{2} \times AB = \dfrac{1}{2} \times CD$

$\texttip{\therefore}{therefore} BP = DQ$

In $\triangle OBP$ and $\triangle ODQ$,

$OD = OQ$ (radii of a circle)

$BP = DQ$ (proved before)

$\angle PBO = \angle QDO$ (proved before)

$\texttip{\therefore}{therefore} \triangle OBP \texttip{\cong}{congruent to} \triangle ODQ$

$\texttip{\therefore}{therefore} OP = OQ$

Since $OP$ and $OQ$ are the distances of the chords $AB$ and $CD$ respectively, from the center, we can say that chords of equal length are equidistant from the center.