Question 1: In an equilateral triangle, prove that the centroid and the center of the circum-circle (circum-center) coincide.2018-12-02_18-36-06

Answer:

Refer to the adjoining diagram.

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' .

Answer:2018-12-02_18-35-53

Refer to the adjoining diagram.

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.2018-12-02_18-35-21

Answer:

Refer to the adjoining diagram.

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 2018-12-02_18-35-34

Answer:

Refer to the adjoining diagram.

\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.2018-12-02_18-34-22

Answer:

Refer to the adjoining diagram.

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 .2018-12-02_18-34-33

Answer:

Refer to the adjoining diagram.

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 .

Answer:2018-12-02_18-34-00

Refer to the adjoining diagram.

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.2018-12-03_8-02-55

Answer:

Refer to the adjoining diagram.

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

Answer:2018-12-03_8-10-28

Refer to the adjoining diagram.

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 .2018-12-02_18-33-15

Answer:

Refer to the adjoining diagram.

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.2018-12-02_18-32-51

Answer:

Refer to the adjoining diagram.

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)

Adding i) and 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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