Congruence of Circle and Arcs

Congruent Circles:

Two circles are said to be congruent if and only if either if them can be super imposed on the other so as to cover it exactly.

This means that two circles are congruent if and only if their radii are equal.  i.e. circles $C(O,r)$ and $C(O, s)$ are congruent if $r = s$.

Congruent Arcs:

Two arcs of a circle are congruent if either of them can be super imposed on the other so as to cover it exactly.

This happens if degree measures of two arcs are the same. Therefore we can say that two arcs of a circle (or of congruent circles) are said to be congruent if and only if they have the same degree measure.

If two arcs $\widehat{PQ}$ and $\widehat{RS}$ are convergent arcs if a circle $C(O,r)$, we write $\widehat{PQ} \cong \widehat{RS}$.

Thus, $\widehat{PQ} \cong \widehat{RS} \Leftrightarrow m(\widehat{PQ}) = m(\widehat{RS}) \Leftrightarrow \angle POQ = \angle ROS$

Important results and their proofs on congruent arcs and chords:

Theorem 1: If two arcs of a circle (or of congruent circles) are congruent, then the corresponding chords are equal.

Given: Two circles $C(O, r)$ and $C(O', r)$. They are congruent hence their radius is equal.

Also $\widehat{PQ} \cong \widehat{RS}$.

Construction: Join $P$ and $Q$ to $O$ in the first circle and $R$ and $S$ to $O'$ in the second circle.

To Prove: $PQ = RS$ (i.e. corresponding chords are equal).

There can be two cases: When $\widehat{PQ}$ and $\widehat{RS}$ are minor arcs (case 1) and When $\widehat{PQ}$ and $\widehat{RS}$ are major arcs (case 2).

Case 1: $\widehat{PQ}$ and $\widehat{RS}$ are minor arcs

Consider $\triangle PQO$ and $\triangle RSO'$

$PO = QO = RO' = SO' = r$ (radius of the two circles are equal)

Also $\angle POQ = \angle RO'S$

(Since $\widehat{PQ} \cong \widehat{RS} \Leftrightarrow m(\widehat{PQ}) = m(\widehat{RS}) \Leftrightarrow \angle POQ = \angle RO'S$ )

Hence $\triangle PQO \cong \triangle RSO'$ (by SAS theorem)

$\Rightarrow PQ = RS$. Hence proved.

Case 2: $\widehat{PQ}$ and $\widehat{RS}$ are major arcs

From case 1, we proved that that

$\widehat{PQ} \cong \widehat{RS}$

$\Rightarrow \widehat{QP} \cong \widehat{SR}$

$\Rightarrow QP = SR$

$\Rightarrow PQ = RS$

Hence $\widehat{PQ} \cong \widehat{RS}$

$\\$

Theorem 2: If two chords of a circle (or of congruent circles) are equal, then the corresponding arcs (major, minor or semi-circular) are congruent.

Given: Chords $PQ$ of circle $C(O,r)$ and $RS$ of circle $C(O',r)$ are equal.

Construction: Join $P$ and $Q$ to $O$ in the first circle and $R$ and $S$ to $O'$ in the second circle. $PQ$ and $RS$ are not diameters.

To Prove: $\widehat{PQ} \cong \widehat{RS}$ where they can be both minor, major or semi-circular arcs.

Proof: There are three possible cases.

Case 1: When $PQ$ and $RS$ are diameters.

In this case, $\widehat{PQ}$ and $\widehat{RS}$ are semi circles of equal radii. Hence they are congruent.

Case 2: When $\widehat{PQ}$ and $\widehat{RS}$ are minor arcs.

Consider $\triangle OPQ$ and $\triangle O'RS$

Given: $PQ = RS$

$PO = O'R = r$

$OQ = O'S = r$

Therefore $\triangle OPQ \cong \triangle O'RS$ (By S.S.S criterion)

$\Rightarrow \angle POQ = \angle RO'S$

$\Rightarrow m(\widehat{PQ}) = m(\widehat{RS})$

$\Rightarrow \widehat{PQ} \cong \widehat{RS}$

Case 3: When $\widehat{PQ}$ and $\widehat{RS}$ are major arcs

$\Rightarrow \widehat{QP}$ and $\widehat{SR}$ are minor arcs

$\Rightarrow QP = SR$

$\Rightarrow m(\widehat{QP}) = m(\widehat{SR})$

$\Rightarrow 360^o - m(\widehat{PQ}) = 360^o - m(\widehat{RS})$

$\Rightarrow m(\widehat{PQ}) = m(\widehat{RS})$

$\Rightarrow \widehat{PQ} \cong \widehat{RS}$

Therefore in all three cases $\widehat{PQ} \cong \widehat{RS}$

$\\$

Theorem 3: The perpendicular from the center of a circle to a chord bisects the chord.

Given: $PQ$ is a chord in Circle $(O,r)$  and $OL \perp PQ$

To Prove: $LP = LQ$

Construction: Join $OP$  and $OQ$

Proof: Consider $\triangle OPL$  and $\triangle OQL$

$OP = OQ = r$  (given)

$OL$  is common

$\angle OLP = \angle OLQ = 90^o$

Therefore $\triangle OPL \cong \triangle OQL$  (By S.S.A criterion)

Therefore $LP = LQ$

Hence $OL$ bisects $PQ$

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Theorem 4 (converse of Theorem 3): If the line joining the center of the circle to the mid point of the chord is perpendicular to the chord.

Given: $M$ is the midpoint of $PQ$

$PQ$ is a chord in circle $C(O,r)$

To Prove: $OM \perp PQ$

Proof: Consider $\triangle OPM$ and $\triangle OMQ$

$OP = OQ = r$

$OM$ is common

$PM = MQ$ (given)

Therefore $\triangle OPM \cong \triangle OMQ$ (By S.S.S criterion)

Therefore $\angle OMP = \angle OMQ$

Since $\angle OMP + \angle OMQ = 180^o$ (linear pairs)

$\Rightarrow 2 \angle OMP = 180^o$

$\Rightarrow \angle OMP = 90^o$

$\Rightarrow \angle OMQ = 90^o$

$\Rightarrow OM \perp OQ$

$\\$

Theorem 4.1: The perpendicular bisectors of two chords of a circle intersect at its center.

Given: $PQ$ and $RS$ are chords in circle $C(O, r)$

$LQ = LP$ ($L$ is the midpoint of $PQ$)

$RM = MS$ ($M$ is midpoint of $RS$)

$O'L \perp PQ$ and $O'M \perp RS$

Let $O'L$ and $O'M$ meet at $O'$

To Prove: $O'$ coincides with $O$

Construction: Join $OL$ and $OM$

Proof: Since $L$ is the mid point of $PQ$

$OL \perp PQ$ (Theorem 4)

$\Rightarrow OL$ is perpendicular bisector of $PQ$

$\Rightarrow OL$ and $O'L$ are perpendicular bisector of $PQ$

$\Rightarrow O'L$ lies on $OL$

Similarly, $M$ is the mid point of $RS$

$\Rightarrow OM \perp RS$

$\Rightarrow OM$ is perpendicular bisector of $RS$ (Theorem 4)

$\Rightarrow OM$ and $O'M$ are perpendicular bisector of $RS$

$\Rightarrow O'M$ lies on $OM$

Therefore, we can say $O'$ (point of intersection of $O'L$ and $O'M$) coincides with $O$ (center of the circle)

Therefore, Perpendicular bisector of $PQ$ and $RS$ will intersect at the center of the circle.

$\\$

Theorem 5: There is one and only one circle passing through three non-collinear points.

Given: $P, Q$ and $R$ are three non-collinear points.

To Prove: One and only one circle passes through $P, Q$ and $R$

Construction: Join $PQ$ and $QR$

Draw perpendicular bisector $AL$ of $PQ$ and also draw perpendicular bisector $BM$ of $RS$

Let $AL$ and $BM$ intersect at $O$

Proof: Since $O$ lies on perpendicular bisector of $PQ$

$\Rightarrow OP = OQ$

Similarly, $OQ = OR$

$\Rightarrow OP = OQ = OR = r$ (radius of the circle)

Now taking $O$ as the center and $r$ as the radius, draw a circle.

You will see that the circle passes through the points $P, Q$ and $R$.

We now need to prove that this is the only circle passing through the points $P, Q$ and $R$.

Now let us assume that there is another circle with center $O' and radius s that goes through point$latex P, Q \$ and $R$.

Then $O'$ will lie on the perpendicular bisector of $AL$ of $PQ$ and $BM$ of $OR$

Since two lines cannot intersect in two points, $O'$ must coincides with $O$

$\Rightarrow r = s$

$\Rightarrow C(O, r) \cong C(O', s)$

Hence only one and one circle can pass through three non-collinear points.

Note 1: In a case where the three points are collinear, then a single circle cannot pass through all of them.

Note 2: This theorem also implies that there will be a unique circle that passes through the vertices of a triangle. This circle is called circum-circle of the triangle and the center of the circle is called circum-center.

Note 3: Infinite number of circles can be drawn through one point.

Note 4: Infinite number of circles can be drawn through two points.

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Theorem 6: If two chords AB and AC of a circle with center O are equal then the center of the circle lies on the angle bisector of $\angle BAC$.

Given: $AB = AC$ are chords of circle $C(O, r)$

To Prove: $O$ lies on the angle bisector of $\angle BAC$

Construction: Join $BC$. Draw bisector of $\angle BAC$

Proof: Consider $\triangle ABM$ and $\triangle ACM$

$AB =AC$ (given)

$AM$ is common

$\angle BAM = \angle CAM$ (By construction)

Therefore $\triangle ABM \cong \triangle ACM$ (By S.A.S criterion)

$\Rightarrow BM = CM$ and $\angle BMA = \angle CAM$

Since $\angle BMA + \angle CMA = 180^o$

$\Rightarrow BMA = \angle CMA = 90^o$

$\Rightarrow AM$ is perpendicular to $BC$ and it bisects $BC$

$\Rightarrow AM$ passes through the center $O$ (By theorem 4)

$\\$

Theorem 7 (converse of Theorem 6): If two chords $AB$ and $AC$ of a circle with center $O$ are such that the center $O$ lies on the bisector of  $\angle BAC$, then $AB = AC$ i.e. chords are equal.

Given: $AB$ and $AC$ are chords of circle $C(O, r)$

$\angle BAO = \angle CAO$

To Prove: $AB = AC$

Construction: Join $OB$ and $OC$

Proof: In $\triangle AOB$

$AO = OB$ (radius)

$\Rightarrow \angle BAO = \angle ABO$ (angles opposite to equal sides are equal)

Similarly, in $\triangle AOC$

$AO = OC$ (radius)

$\Rightarrow \angle OAC = \angle OCA$ (angles opposite to equal sides are equal)

But $\angle BAO = \angle CAO$

$\Rightarrow \angle BAO = \angle ABO = \angle OAC = \angle OCA$

Consider $\triangle OAB$ and $\triangle OAC$

$\angle OBA = \angle OCA$

$\angle OAB = \angle OAC$

$OA$ is common

Therefore $\triangle OAB \cong \triangle OAC$

$\Rightarrow AB = AC$ (since corresponding parts of congruent triangles are equal)

$\\$

Theorem 8: If two circles intersect in two points, then the line through the centers of the circles is perpendicular bisector of the common chord.

Given: Two circles $C(O, r)$ and $C(O',s)$

They intersect at $A$ and $B$

Construction: Join $OA$ and $OB$. Also join $O'A$ and $O'B$

To Prove: $OO'$ is perpendicular bisector of $AB$

Proof: Consider $\triangle OAO'$ and $\triangle OBO'$

$OA = OB = r$

$O'A = O'B = s$

$OO'$ is common

Therefore $\triangle OAO' \cong \triangle OBO'$ (By S.S.S criterion)

$\Rightarrow \angle AOO' = \angle BOO'$

$\Rightarrow \angle AOM = \angle BOM$

Let $OO'$ and $AB$ intersect at $M$

Consider $\triangle AOM$ and $\triangle BOM$

$OA = OB = r$

$\angle AOM = \angle BOM$

$OM$ is common

Therefore $\triangle AOM \cong \triangle BOM$ (By S.A.S criterion)

$\Rightarrow AM = BM$ and $\angle AMO = \angle BMO$

We know $\angle AMO + \angle BMO = 180^o$

$\Rightarrow 2 \angle AMO = 180^o$

$\Rightarrow AMO = 90^o$

Thus $AM = BM$ and $\angle AMO = \angle BMO = 90^o$

Hence $OO'$ is perpendicular bisector of $AB$