Question 1: In an equilateral triangle, prove that the centroid and the center of the circum-circle (circum-center) coincide.

Given: $\triangle ABC$ is an equilateral triangle. $D, E$ and $F$ are mid points of $BC, AC$ and $AB$ respectively.

To Prove: Centroid and Circum-center are coincident.

Construction: Draw medians $AD, BE$ and $CF$

Proof: Consider $\triangle BEC$ and $\triangle BFC$

$\angle FBC = \angle ECB = 60^o$ (given)

BC is common

Since $F$ and $E$ are mid points of $AB$ and $AC$ respectively, and $AB = AC$ because $\triangle ABC$ is equilateral we have $BF = CE$

Therefore $\triangle BEC \cong \triangle BFC$ (By S.A.S criterion)

$\Rightarrow BE = CF$ … … … … … i)

Similarly, $\triangle CAF \cong \triangle CAD$

$\Rightarrow CF = AD$ … … … … … ii)

From i) and ii) we get

$BE = CF = AD$

$\Rightarrow$ $\frac{2}{3}$ $BE =$ $\frac{2}{3}$ $CF =$ $\frac{2}{3}$ $AD$

$\Rightarrow BO = CO = AO$

$\Rightarrow O$ is equidistant from the vertices

$\Rightarrow O$ is the circum-center of $\triangle ABC$

Question 2: Two circles whose centers are $O$ and $O'$ intersect at $P$. Through $P$, a line $l \parallel OO'$ intersecting the circles at $C$ and $D$ is drawn. Prove that $CD = 2 OO'$.

Construction: Draw $OA \perp l$ and $OB \perp l$

Proof: $OA \perp l \Rightarrow OA \perp CP$

$\Rightarrow CA = AP$

$\Rightarrow CP = 2 AP$ … … … … … i)

(perpendicular drawn from the center of the circle bisect the chord)

Similarly, $O'B \perp l \Rightarrow O'B \perp PD$

$\Rightarrow PB = BD$

$\Rightarrow PD = 2 PB$ … … … … … ii)

Therefore $CD = CP + PD$

$\Rightarrow CD = 2 AP + 2 PB$

$CD = 2 (AP + PB) = 2 AB = 2 OO'$

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Question 3: Prove that the line joining the mid points of two parallel chords of a circle passes through the center.

Construction: Join $OP$ and $OQ$. Draw $OX \parallel AB$ and $CD$

Proof:  Since $P$ is the mid point of $AB$

$\Rightarrow OP \perp AB$ (Theorem 4)

$\Rightarrow \angle OPB = 90^o$

Since $AB \parallel OX$

Therefore $\angle OPB + \angle POX = 180^o$

$\Rightarrow \angle POX = 180^o - 90^o = 90^o$

Similarly, $OQ \perp CD$

$\Rightarrow \angle OQD = 90^o$ (Theorem 4)

Since $CD \parallel OX$

Therefore $\angle XOQ + \angle OQD = 180^o$

$\Rightarrow \angle XOQ = 180^o - 90^o = 90^o$

Hence $\angle POX + \angle XOQ = 90^o + 90^o = 180^o$

Hence $PQ$ is a straight line.

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Question 4: In the adjoining figure $\widehat{AB} \cong \widehat{CD}$, Prove that $\angle A = \angle B$

$\widehat{AB} \cong \widehat{CD}$

$\angle AOB = \angle COD$

Since congruent arcs of a circle subtend equal angles at the center

Therefore $\angle AOB + \angle BOC = \angle COD + \angle BOC$

$\Rightarrow \angle AOC = \angle BOD$

Consider $\triangle AOC$ and $\triangle BOD$

$AO = OB$ (radius)

$OC = OD$ (radius)

$\angle AOC = \angle BOD$

Therefore $\triangle AOC \cong \triangle BOD$ (By S.A.S criterion)

Hence $\angle A = \angle B$

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Question 5: If two chords of a circle are equally inclined to the diameter through their point of intersection, prove that the chords are equal.

Given: $\angle OAL = \angle OAM$

Construction: Draw $OL \perp AB$ and $OM \perp AC$

To Prove: $AB = AC$

Proof: Consider $\triangle AOL$ and $\triangle AOM$

$AO$ is common

$\angle OAL = \angle OAM$  (given)

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

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

Hence $OL = OM$

Since equidistant chords are equal, $AB = AC$

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Question 6: In adjoining figure, $O$ is the center of a circle and $PO$ bisects $\angle APD$. Prove that $AB = CD$.

To Prove: $AB = CD$

Construction: Draw $OE \perp AB$ and $OF \perp CD$

Proof: Consider $\triangle OFP$ and $\triangle OEP$

$OP$ is common

$\angle OFP = \angle OEP = 90^o$

$\angle OPF = \angle OPE$ ($OP$ bisects $\angle APD$ – given)

Therefore $\triangle OFP \cong \triangle OEP$

$\Rightarrow OE = OF$

$\Rightarrow AB$ and $CD$ are equidistant

Hence $AB = CD$ (equidistant chords in a circle are equal)

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Question 7: Two equal chords $AB$ and $CD$ of a circle with center $O$, when produced meet at a point $E$ as shown in the adjoining diagram. Prove that $BE = DE$ and $AE = CE$.

Given: $AB = CD$

Construction: Draw $OL \perp AB$ and $OM \perp CD$

To Prove: $BE = DE$ and $AE = CF$

Proof: Consider $\triangle OLE$ and $\triangle OME$

$OE$ is common

$\angle OLE = \angle OME = 90^o$

Since $AB = CD$

$\Rightarrow$ they are equidistant

Therefore $OL = OM$

Hence $\triangle OLE \cong \triangle OME$ (By S.A.S criterion)

Therefore $LE = ME$ … … … … … i)

We know $AB = CD \Rightarrow \frac{1}{2} AB = \frac{1}{2} CD \Rightarrow BL = DM$ … … … … … ii)

i) – ii) we get

$LE - BL = ME - DM \Rightarrow BE = DE$.

Since $AB = CD$ and $BE = DE$,

$AB + BE = CD + DE$

$\Rightarrow AE = CE$. Hence proved

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Question 8: Prove that the line joining the mid points of two equal chords of a circle subtend equal angles with the chords.

Given: $AB = CD, L$ and $M$ are mid points of $AB$ and $CD$ respectively.

To Prove: $\angle ALM = \angle CML$ and $\angle BLM = \angle DML$

Construction: Draw $OL \perp AB$ and $OM \perp CD$

Proof: $OL = OM$ (equal chords in a circle are equidistant)

Therefore $\angle OLM = \angle OML$

$\angle OLM + 90^o = \angle OML + 90^o$

$\angle BLM = \angle DML$

Similarly, $90^o - \angle OLM = 90^o - \angle OML$

$\Rightarrow \angle ALM = \angle LMC$

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Question 9: In adjoining figure, $L$ and $M$ are mid points of two equal chords $AB$ and $CD$ of a circle with center $O$. Prove that i) $\angle OLM = \angle OML$ ii) $\angle ALM = \angle CML$

Given: $AB = CD, OL \perp AB$ and $OM \perp CD$

Proof: Since equal chords are equidistant from the center, $OL = OM$

In $\triangle OLM$

$\angle OLM = \angle OML$ (angle opposite equal sides of a triangle are equal)

$90^o - \angle OLM = 90 - \angle OML$

$\Rightarrow \angle ALM = \angle CML$

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Question 10: $PQ$ and $RQ$ are chords of a circle equidistant from the center. Prove that the diameter passing through $Q$ bisects $\angle PQR$ and $\angle PSR$.

Given: $PQ$ and $RS$ are equidistant from $O$

To Prove:

$\angle PQS = \angle RQS$ ($QS$ bisects $\angle PQR$)

$\angle PSQ = \angle RSQ$ ($QS$ bisects $\angle PSR$)

Construction: Join $PS$ and $RS$

Proof: Equidistant chords are equal

$\Rightarrow PQ = QR$

Consider $\triangle PQS$ and $\triangle RQS$

$QS$ is common

$QP = QR$

$\angle QPS = \angle QRS = 90^o$ (angles subtended by diameter)

$\Rightarrow \triangle PQS \cong \triangle RQS$ (By. S.A.S criterion)

$\Rightarrow \angle PQS = \angle RQS$ and $\angle PSQ = \angle RSQ$. Hence proved.

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Question 11: If two chords of a circle bisect each other, show that they must be diameters.

Construction: Join $AC, BD, AD$ and $BC$.

To Prove: $AB$ and $CD$ are diameters

Proof: Consider $\triangle AOC$ and $\triangle BOD$

$OC = OD$ (Mid point of $CD$)

$OB = OA$ (Mid point of $AB$)

$\angle AOC = \angle BOD$ (vertically opposite angles)

$\triangle AOC \cong \triangle BOD$  (By S.A.S criterion)

$\Rightarrow AC = BD$

$\Rightarrow \widehat{AC} \cong \widehat{BD}$ … … … … … i)

Now consider $\triangle AOD$ and $\triangle BOC$

$OA = OB$ (Mid point of $AB$)

$OC = OD$ (Mid point of $CD$)

$\angle AOD = \angle COD$ (vertically opposite angles)

$\triangle AOD \cong \triangle BOC$

$\Rightarrow AD = BC$

$\Rightarrow \widehat{AD} \cong \widehat{BC}$ … … … … … ii)

$\widehat{AC} + \widehat{AD} = \widehat{BD} + \widehat{BC}$

$\widehat{CAD} \cong \widehat{CBD}$

$\Rightarrow CD$ divides the circle in 2 half’s

$\Rightarrow CD$ is a diameter

Similarly, $\widehat{AC} + \widehat{BC} = \widehat{BD} + \widehat{AD}$

$\widehat{ACB} \cong \widehat{BDA}$

Therefore $AB$ is diameter

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