Results of Equal Chords

Theorem 1: Equal chords are equidistant from the center.

Given: $OL = OM$

Construction: Join $OA$ and $OC$

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Since $OL \perp AB \Rightarrow AL = \frac{1}{2} AB$

Similarly, $OM \perp CD \Rightarrow CM = \frac{1}{2} CD$

Given $AB = CD$

$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$

$\Rightarrow AL = CM$

Now consider $\triangle OMC$ and $\triangle OLA$

$AL = CM$ (proved above)

$OA = OC$ (radius)

$\angle OLA = \angle OMC = 90^o$

Therefore $\triangle OMC \cong \triangle OLA$ (By RHS criterion)

$\Rightarrow OL = OM$. Hence proved that equal chords are equidistant from the center.

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Theorem 2 (Converse of Theorem 1): Chords of a circle which are equidistant from the center are equal.

Given: Two chords $AB$ and $CD$ in circle $C(O, r)$

Also $OL = OM$ where $OL \perp AB$ and $OM \perp CD$

To Prove: $AB = CD$

Construction: Join $OA$ and $OC$

Proof: $OL \perp AB$ (given)

We know that the perpendicular from the center of a circle to a chord, bisects the chord.

$\Rightarrow AL = BL$

$\Rightarrow AL = \frac{1}{2} AB$

Similarly $OM \perp CD$

$\Rightarrow CM = DM$

$\Rightarrow CM = \frac{1}{2} CD$

Consider $\triangle OAL$ and $\triangle OCM$

$OA = OC$ (radius)

$\angle OLA = \angle OMC = 90^o$

$OL = OM$ (given)

Therefore $\triangle OAL \cong \triangle OCM$ (By S.A.S criterion)

$\Rightarrow AL = CM$

$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$

$\Rightarrow AB = CD$

Hence proved that chords of a circle which are equidistant from the center are equal.

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Theorem 3: Equal chords of concurrent circles are equidistant from the corresponding centers.

Given: Two concurrent circles $C(O,r)$ and $C(O',r)$ and $AB = CD$

Construction: Join $OA$ and $O'C$

To Prove: $OL = O'M$ where $OL \perp AB$ and $O'M \perp CD$

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

$AL = BL \Rightarrow AL = \frac{1}{2} AB$

Similarly, $CM = DM \Rightarrow CM = \frac{1}{2} CD$

Now $AB = CD$ (given)

$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$

$\Rightarrow AL = CM$

Consider $\triangle OAL$ and $\triangle O'CM$

$OA = O'C$ (congruent circles with radius $r$)

$\angle OLA = \angle O'MC = 90^o$

$AL = CM$ (proved above)

Therefore $\triangle OAL \cong \triangle OC'M$ (By RHS criterion)

Therefore $OL = O'M$ i.e. $AB$ and $CD$ are equidistant from $O$ and $O'$ respectively.

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Theorem 4 (Converse of Theorem 3): Chords of congruent circles which are equidistant from their corresponding centers are equal.

Given: Two circles $C(O,r)$ and $C(O',r)$. $AB$ and $CD$ are chords such that $OL = O'M$ . $OL \perp AB$ and $O'M \perp CD$

To Prove: $AB = CD$

Construction: Join $OA$ and $O'C$

Proof: Consider $\triangle OLA$ and $\triangle O'MC$

$OL = O'M$ (given)

$OA = O'C$ (equal radius)

$\angle OLA = \angle O'MC = 90^o$

Therefore $\triangle OLA \cong \triangle O'MC$ (By RHS criterion)

$\Rightarrow AL = CM$

$\Rightarrow 2 AL = 2 CM$

We know that the perpendicular from the center of a circle to a chord, bisects the chord.

$\Rightarrow AB = CD$

Hence proved that chords of congruent circles which are equidistant from their corresponding centers are equal.

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Theorem 5: Equal chords of a circle subtends equal angles at the center.

Given: $AB$ and $CD$ are two equal chords in circle $C(O,r)$

To Prove: $\angle AOB = \angle COD$

Proof: Consider $\triangle AOB$ and $\triangle COD$

$OA = OD$ (radius)

$OB = OC$ (radius)

$AB = CD$ (given)

Therefore $\triangle AOB \cong \triangle COD$

$\Rightarrow \angle AOB = \angle COD$.

Hence proved that equal chords of a circle subtend equal angles at the center.

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Theorem 6: (converse of Theorem 5) If angles subtended by two chords of a circle at the center are equal, then the chords are equal.

Given: $AB$ and $CD$ are chords of a circle $C(O,r)$ such that $\angle AOB = \angle COD$

To Prove: $AB = CD$

Proof: Consider $\triangle AOB$ and $\triangle COD$

$OA = OD$ (radius)

$OB = OC$ (radius)

$\angle AOB = \angle COD$ (given)

Therefore $\triangle AOB \cong \triangle COD$ (By S.A.S criterion)

Therefore $AB = CD$ (corresponding sides off congruent triangles are equal)

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Theorem 7: Equal chords of congruent triangles subtend equal angles at the center.

Given: Two concurrent circles $C(O, r)$ and $C(O', r)$. Chords $AB = CD$

To Prove: $\angle AOB = \angle CO'D$

Proof: Consider $\triangle AOB$ and $\triangle CO'D$

$OA = O'C$ (radius)

$OB = O'D$ (radius)

$AB = CD$ (given)

Therefore $\triangle AOB \cong \triangle CO'D$ (By S.S.S criterion)

Hence $\angle AOB = \angle CO'D$ (corresponding angles of congruent triangles are equal).

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Theorem 8 (converse of Theorem 7): If the angles subtended by the two chords of congruent circles at the corresponding centers of circles are equal, the chords are equal.

Given: Two congruent circles $C(O,r)$ and $C(O',r)$. AB and CD are chords such that $\angle AOB = \angle CO'D$

To Prove: $AB = CD$

Proof: Consider $\triangle AOB$ and $\triangle CO'D$

$OA = O'C$ (radius)

$OB = O'D$ (radius)

$\angle AOB = \angle CO'D$ (given)

Therefore $\triangle AOB \cong \triangle CO'D$ (By S.A.S criterion)

$\Rightarrow AB = CD$

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Theorem 9: Of any two chords of a circle, the chord which is larger is nearer to the circle.

Given: Two chords $AB$ and $CD$ of a circle with center $O$ such that $AB > CD$

Construction: Join $OA$ and $OC$

To Prove: $OL < OM$ or $OM > OL$

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

$OL \perp AB \Rightarrow AL = \frac{1}{2} AB$

And $OM \perp CD \Rightarrow CM = \frac{1}{2} CD$

In right triangles $\triangle OAL$ and $\triangle OCM$, we have

$OA^2 = OL^2 + AL^2$

And $OC^2 = OM^2 + CM^2$

Since $OA = OC$ (radius)

$OL^2 + AL^2 = OM^2 + CM^2$

Given $AB > CD$

$\Rightarrow \frac{1}{2} AB > \frac{1}{2} CD$

$\Rightarrow AL > CM$

$\Rightarrow AL^2 > CM^2$ (Adding $OL^2$ to both sides)

Therefore $OL^2 + AL^2 > OL^2 + CM^2$

$\Rightarrow OM^2 + CM^2 > OL^2 + CM^2$

$\Rightarrow OM^2 > OL^2$

$\Rightarrow OM > OL.$

Hence proved that of any two chords of a circle, the chord which is larger is nearer to the circle.

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Theorem 10 (converse of Theorem 9): Of any two chords of a circle, the chord which is nearer is larger.

Given: Two chords $AB$ and $CD$ of a circle $C(O, r)$ such that $OL < OM$, where $OL$ and $OM$ are perpendicular from $O$ on $AB$ and $CD$ respectively.

To Prove: $AB > CD$

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Therefore $AL = \frac{1}{2} AB \ \ \ \& \ \ \ \ CM = \frac{1}{2} CD$

In right triangles $\triangle OAL$ and $\triangle OCM$ we have

$OA^2 = OL^2 + AL^2$

And $OC^2 = OM^2 + CM^2$

$\Rightarrow AL^2 = OA^2 - OL^2$ … … … … … i)

And $CM^2 = OC^2 - OM^2$ … … … … … ii)

Now $OL < OM$

$\Rightarrow OL^2 < OM^2$

$\Rightarrow = OL^2 > - OM^2$ (Adding $OA^2$ on both sides)

$OA^2 - OL^2 > OA^2 - OM^2$

$AL^2 > OA^2 - OM^2$

$AL^2 > OC^2 - OM^2$

Since $OA^2 = OC^2$ (radius of the circle)

$AL^2 > CM^2$

$\Rightarrow AL > CM$

$\Rightarrow 2AL > 2 CM$

$\Rightarrow AB > CD$. Hence proved that of any two chords of a circle, the chord which is nearer is larger.