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Circles - Class 9 : Notes


 

(1) A circle is the collection of those points in a plane that are at a given constant distance from a fixed-point in the plane. The fixed point is called the centre and the given constant distance is called the radius of the circle.
A Circle with centre O and radius r usually denoted by C(O,r). Thus, in set theoretic notations, we write C(O,r) = \left\{ {X:OX = r} \right\}

(2) A point P lies inside or on or outside the circle C(O,r) according as OP < r or OP = r or OP > r.

(3) The collection of all points lying inside and on the circle C(O,r) is called a circular disc with centre O and radius r.
The set of all points lying inside and on the circle is called a Circular Disc. It is also known as the circular region.

(4) Circles having the same center and different radii are said to be concentric circles.
When two or more circles have the same center but have different radii, they are called as concentric circles, that is, circles with common center.

(5) A continuous piece of a circle is called an arc of the circle.
For Example: Consider circle C (O, r). Let {P_1},{P_2},{P_3},{P_4},{P_5},{P_6} be point on the circle. Then, the pieces {P_1},{P_2},{P_3},{P_4},{P_5},{P_6},{P_1},{P_2} etc. are all arcs of the circle C\left({O,r} \right).

(6) Prove that If two arcs of circle are congruent,  then corresponding chords are equal.

Given: Arc PQ of a Circle C\left({O,r} \right) and arc RS of another circle C(O',r) such that PQ \cong RS
To Prove: PQ = RS
Construction: Draw Line segment OP, OQ, O'R and O'S.Proof:
Case-I When arc(PQ) and arc(RS) are minor Arcs

In triangle OPQ and O'RS, We have
OP = OQ = O'R = O'S = r                   [Equal radii of two circles]
\angle POQ = \angle RO'S                 arc(PQ) \cong arc(RS) \Rightarrow m(arc(PQ)) \cong m(arc(RS)) \Rightarrow \angle POQ = \angle RO'S
So by SAS Criterion of congruence, we have
\Delta POQ \cong \Delta R{O^{'}}S
 \Rightarrow PQ = RS
Case-II When arc(PQ) and arc(RS) are major arcs.
If arc(PQ), arc(RS) are major arcs, then arc(QP) and arc(SR) are Minor arcs.
So arc(PQ) \cong arc(RS)
 \Rightarrow  arc(QP) \cong arc(SR)
 \Rightarrow QP = SR
 \Rightarrow PQ = RS
Hence, PQ \cong RS  \Rightarrow PQ = RS

(7) Prove that If two chords of a circle are equal, then their corresponding arcs are congruent.

Given: Equal chords, PQ of a circle C\left( {O,r} \right)  and RS of congruent circle C\left( {{O^{'}},r} \right)
To Prove: arc(PQ) \cong arc(RS), where both arc(PQ) and arc(RS) are minor, major or semi-circular arcs.
Construction: If PQ,RS are not diameters, draw line segments OP, OQ, {O^{'}}R and {O^{'}}S.Proof:
Case I: when arc(PQ) and arc(RS) are diameters

In this case, PQ and RS are semi circle of equal radii, hence they are congruent.
Case II: When arc(PQ) and arc(RS) are Minor arcs.
In triangles POQ and R{O^{'}}S, we have
PQ = RS
OP = {O^{'}}R = r and OQ = {O^{'}}S = r
So by SSS-criterion of congruence, we have
\Delta POQ \cong \Delta R{O^{'}}S
 \Rightarrow             \angle POQ = \angle R{O^{'}}S
 \Rightarrow           m(arc(PQ))=m(arc(RS))
 \Rightarrow         arc(PQ) \cong arc(RS)
Case III: When arc(PQ) and arc(RS) are major arcs
In this case, arc(QP) And arc(SR)will be minor arcs.
PQ = RS
 \Rightarrow            QP = SR
 \Rightarrow           m(arc(QP))=m(arc(SR))
 \Rightarrow           {360^ \circ } - m(arc(PQ)) - {360^ \circ } - m(arc(RS))
 \Rightarrow           m(arc(PQ)) - m(arc(RS))
 \Rightarrow           arc(PQ) \cong arc(RS)
Hence, in all the three cases, we have arc(PQ) \cong arc(RS)

(8) Prove that The perpendicular from the centre of a circle to a chord bisects the chord.

Given: A Chord PQ of a circle C(O,r) and perpendicular OL to the chord PQ.
To Prove: LP = LQ
Construction: Join OP and OQProof: In Triangles PLO and QLO, we have
OP = OQ = r                               [Radii of the same circle]
OL = OL                                       [Common]
And,      \angle OLP = \angle OLQ     [Each equal to {90^ \circ }]
So, by RHS-criterion of congruence, we have
\Delta PLO \cong \Delta QLO
 \Rightarrow             PL = LQ

(9) Prove that The line segment joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.

Given: A Chord PQ OF a circle C(O,r) with mid-point M.
To Prove: OM \bot PQ
Construction: Join OP and OQProof: In triangles OPM and OQM, we have
OP = OQ                                     [Radii of the same circle]
PM = MQ                                   [M is mid-point of PQ]
OM = OM
So, by SSC- criterion of congruence, we have
\Delta OPM \cong \Delta OQM
 \Rightarrow             \angle OMP = \angle OMQ
But , \angle OMP + \angle OMQ = {180^ \circ }                                          [Linear pair]
 \Rightarrow             \angle OMP + \angle OMP = {180^ \circ }  [\angle OMP = \angle OMQ]
 \Rightarrow             2\angle OMP = {180^ \circ }
 \Rightarrow             \angle OMP = {90^ \circ }

(10) Prove that There is one and only circle passing through three given points.

Given: Three non-collinear points P,Q and R.
To Prove: There is one and only one circle passing through P,Q and R.
Construction: Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P,Q and R. are not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel.
Let AL And BM intersect at O. Join OP,OQ and OR.

Proof: Since O lies on the perpendicular bisector of PQ.
Therefore,
OP = OQ
Again, O Lies on the perpendicular bisector of QR.
Therefore,
OQ = OR
Thus,     OP = OQ = OR = r    (say)
Taking O as the centre draw a circle of radius s. Clearly, C(O,s) passes through P, Q and R. This proves that there is a circle passing the points P,Q and R.
We shall now prove that this is the only circle passing through P,Q and R.
If possible, let there be another circle with centre {O^{'}} and radius r, passing through the points P,Q and R. Then, {O^{'}} will lie on the perpendicular bisectors AL of PQ and BM of QR.
Since two lines cannot intersect at more than one point, so {O^{'}} must coincide with O. Since OP=r, {{O^{'}}P}=s and O and {O^{'}} coincide, we must have
r = s
 \Rightarrow            C(O,r) = C({O^{'}},s)
Hence, there is one and only one circle passing through three non-collinear points P,Q and R.

(11) Prove that If two circles intersect in two points, then the through the centre is perpendicular to the common chord.

Given: Two circles C(O,r) and C({O^{'}},s) intersecting at points A and B.
To Prove: O{O^{'}} is perpendicular bisector of AB.
Construction: Draw line segments
OA,OB,{O^{'}}A and {O^{'}}BProof: In triangles OA{O^{'}} and OB{O^{'}},  we have
OA = OB = r
{O^{'}}A = {O^{'}}B = s
And,                O{O^{'}}=O{O^{'}}
So, by SSS-criterion of congruence, we have
\Delta OAO \cong \Delta OB{O^{'}}
 \Rightarrow        \angle AO{O^{'}} = \angle BO{O^{'}}
 \Rightarrow        \angle AOM= \angle BOM         [\angle AO{O^{'}} = \angle AOM and \angle BOM = \angle BO{O^{'}}]
Let M be the point of intersection of AB and O{O^{'}}
In triangles AOM and BOM, we have
OA = OB = r
 \Rightarrow        \angle AO{O^{'}} = \angle BO{O^{'}}
 \Rightarrow        \angle AOM = \angle BOM        [\angle AO{O^{'}} = \angle AOM and \angle BOM = \angle BO{O^{'}}]
Let M be the point of intersection of AB and O{O^{'}}
In triangles AOM and BOM, we have
OA = OB = r
\angle AOM = \angle BOM
And                 OM=OM
So, by SAS-criterion of congruence, we have
\Delta AOM \cong \Delta BOM
 \Rightarrow        AM=BM and \angle AMO= \angle BOM
But,                             \angle AOM + \angle BMO = {180^ \circ }
2\angle AOM = {180^ \circ }
 \Rightarrow        \angle AOM = {90^ \circ }
Thus,                           AM=BM and \angle AOM = \angle BMO = {90^ \circ }
Hence, O{O^{'}} is the perpendicular bisector of AB.

(12) Prove that Equal chords of a circle subtend equal angle at the centre.

Given: Two Chord AB and CD of circle C(O,r) such that AB = CDand OL \bot AB and OM \bot CD
To Prove: Chord AB and CD are equidistant from the centre O i.e OL=OM.
Construction: Join OA and OC.Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
OL \bot AB  \Rightarrow AL = {1 \over 2}AB..........(i)
And,                OM\bot CD  \Rightarrow CM= {1 \over 2}CD..........(ii)
But,                 AB=CD
 \Rightarrow        {1 \over 2}AB = {1 \over 2}CD
 \Rightarrow        AL = CM                                [Using (i) and (ii) ]….......(iii)
Now, in right triangles OAL and OCM, we have
OA = OC                                           [Equal to radius of the circle]
AL = CM                                            [From equation (iii)]
And,                \angle ALO = \angle CMO         [Each equal to {90^ \circ }]
So by RHS criterion of convergence, we have
\Delta OAL \cong \Delta OCM
 \Rightarrow        OL= OM
Hence, equal chord of a circle are equidistant from the centre.

(13) Prove that Chords of a circle which are equidistant from the centre are equal.

Given: Two Chords AB and CD of a circle C(O,r) which are equidistant from its centre i.e. OL = OM, where OL \bot AB and OM \bot CD.
To Prove: Chords are Equal i.e. AB = CD
Construction: Join OA and OCProof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
OL \bot AB
 \Rightarrow        AL = BL
 \Rightarrow        AL = {1 \over 2}AB
And,                            OM \bot CD
 \Rightarrow        CM = DM
 \Rightarrow        CM = {1 \over 2}CD
In triangles OAL and OCM, we have
OA = OC                        [Each equal to radius of the given Circle]
\angle OLA = \angle OMC         [Each equal to {90^ \circ }]
And,                OL = OM                                            [Given]
So, by RHS, criterion of convergence, we have
\Delta OAL \cong \Delta OCM
 \Rightarrow        AL = CM
 \Rightarrow        {1 \over 2}AL = {1 \over 2}AB
 \Rightarrow        AB = CD
Hence, the chords of a circle which are equidistant from the centre are equal.

(14) Prove that Equal chords of a circle subtend equal angle at the centre.

Given: A circle C(O,r) and its two equal chords AB and CD.
To Prove: \angle AOB = \angle CODProof:In triangles AOB and COD, we have
AB = CD                                                [Given]
OA= OC                                     [Each equal to r]
OB = OD                                                [Each equal to r]
So, by SSC-criterion of Congruence, we have
\Delta AOB \cong \Delta COD
 \Rightarrow        \angle AOB = \angle COD

(15) Prove that If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.

Given: Two Chord AB and CD of a circle C(O,r) such that \angle AOB = \angle COD
To Prove: AB = CDProof: In triangles AOB and COD, we have
OA= OC                                     [Each equal to r]
\angle AOB = \angle COD         [Given]
OB= OD                                     [Each equal to r]
So, by SAS-criterion of congruence, we have
\Delta AOB \cong \Delta COD
 \Rightarrow        AB=CD

(16) Prove that Of any two chords of a circle, the larger chord is nearer to the centre.

Given: Two Chord AB and CD of a circle with Centre O such that AB > CD
To Prove: Chord AB is nearer to the centre of the circle i.e. OL < OM, where OL and OM are perpendiculars from O to AB and CD respectively
Construction: Join OA and OC.Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
OL \bot AB  \Rightarrow AL = {1 \over 2}AB
And,                OM \bot CD  \Rightarrow CM = {1 \over 2}CD
In right triangles OAL and OCM, we have
O{A^2} = O{L^2} + A{L^2}
And,                O{C^2} = O{M^2} + C{M^2}
 \Rightarrow         O{L^2} + A{L^2}= O{M^2}+ C{M^2} .. (i)  [OA = OC \Rightarrow O{A^2} = O{C^2}]
Now,                           AB > CD
 \Rightarrow        {1 \over 2}AB > {1 \over 2}CD
 \Rightarrow        AL > CM
 \Rightarrow        A{L^2} > C{M^2}
 \Rightarrow         O{L^2} +A{L^2} > O{L^2} + C{M^2}         [Adding O{L^2} on both sides]
 \Rightarrow         O{M^2} +C{M^2} > O{L^2} + C{M^2}       [using equation (i)]
 \Rightarrow         O{M^2} > O{L^2}
 \Rightarrow        OM > OL
 \Rightarrow        OL < OM
Hence, AB is nearer to the centre than CD.

(17) Prove that Of any two chords of a circle, the chord nearer to the centre is larger.

Given: Two Chord AB and CD of a circle C(O,r) such that OL < OM, where OL and OM are perpendiculars From O on AB and CD respectively.
To Prove: AB > CD
Construction: Join OA and OC.Proof: Since the perpendicular From the Centre of a circle to a chord bisects the chord.
AL = {1 \over 2}AB and CM = {1 \over 2}CD
In right triangles OAL and OCM, we have
O{A^2} = O{L^2} + A{L^2} and, O{C^2} = O{M^2} + C{M^2}
 \Rightarrow         A{L^2} = O{A^2}- O{L^2}....... (i)
And,                            C{M^2} = O{C^2} - O{M^2}.......(ii)
Now,                           OL < OM
 \Rightarrow        O{L^2} < O{M^2}
 \Rightarrow         - O{L^2} > - O{M^2}
 \Rightarrow        O{A^2} - O{L^2} > O{A^2} - O{M^2}         [adding O{A^2} on both sides]
 \Rightarrow        O{A^2} - O{L^2} > O{C^2} - O{M^2}          [O{A^2} = O{C^2}]
 \Rightarrow        A{L^2} > C{M^2}
 \Rightarrow        AL > CM
 \Rightarrow        2AL > 2CM
 \Rightarrow        AB > CD

(18) Prove that The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given: An arc PQ of a circle C(O,r) and a point R on the remaining part of the circle i.e. arc QP.
To Prove: \angle POQ = 2\angle PRQ
Construction: join RO and produce it to a point M outside the circle.

Proof: We shall consider the following three different cases:
Case I: when arc(PQ) is a minor arc.
We know that an exterior angle of a triangle is equal to the sum of the interior oppsite angles.
In \Delta POQ, \angle POM is the exterior angle.
\angle POM = \angle OPR + \angle ORP
 \Rightarrow \angle POM = \angle ORP + \angle ORP     [OP = OR = r, \angle OPR = \angle ORP] \Rightarrow \angle POM = 2\angle ORP .....................(i)
In \Delta QOR, \angle QOM is the exterior angle.
\angle QOM = \angle OQR + \angle ORQ
 \Rightarrow \angle QOM = \angle OQP + \angle ORQ   [OQ = OR = r, \angle ORQ = \angle OQR]
 \Rightarrow \angle QOM = 2\angle ORQ ................(ii)
Adding equation (i) and (ii), we get
\angle POM + \angle QOM = 2\angle ORP + 2\angle ORP
 \Rightarrow \angle POM + \angle QOM = 2(\angle ORP + \angle ORP)
 \Rightarrow   \angle POM = 2\angle PRQ
Case II: when arc(PQ) is a semi-circle
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles.
In \Delta POQ, we have
\angle POM = \angle OPR + \angle ORP
 \Rightarrow \angle POM = \angle ORP + \angle ORP     [OP = OR = r, \angle OPR = \angle ORP]
 \Rightarrow \angle POM = 2\angle ORP .....................(iii)
In \Delta QOR, We have
\angle QOM = \angle ORQ + \angle OQR
 \Rightarrow \angle QOM = \angle ORQ + \angle ORQ   [OQ = OR = r, \angle ORQ = \angle OQR]
 \Rightarrow \angle QOM = 2\angle ORQ ................(iv)
Adding equations (iii) and (iv), we get
\angle POM + \angle QOM = 2(\angle ORP + \angle ORQ)
\angle POQ = 2 \angle PRQ
Case III: When arc(PQ) is a major arc.
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles
In \Delta POR, we have
\angle POM = \angle ORP + \angle ORP           [OP = OR = r, \angle OPR = \angle ORP]
 \Rightarrow \angle POM = 2\angle ORP ...................(v)
In \Delta QOR, we have
\angle QOM = \angle ORQ + \angle OQR
 \Rightarrow \angle QOM = 2\angle ORQ ...................(vi)
Adding equations (v) and (vi), we get
\angle POM + \angle QOM = 2(\angle ORP + \angle ORP)
 \Rightarrow Reflex \angle POQ = 2\angle PRQ

(19)Prove that Angles in the same segment of a circle are equal.

Given: A circle C(O,r), an arc PQ and two angles \angle PRQ and \angle PSQ in the same segment of  the circle.
To Prove: \angle PRQ= \angle PSQ
Construction: Join OP and OQProof: we know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point in the remaining part of the circle. So we have
\angle POQ = 2\angle PRQ and \angle POQ = 2\angle PSQ
 \Rightarrow 2\angle PRQ = 2\angle PSQ
 \Rightarrow \angle PRQ = \angle PSQ
We have
Reflex \angle POQ = 2\angle PRQ and \angle POQ = 2\angle PSQ
 \Rightarrow 2\angle PRQ = 2\angle PSQ
 \Rightarrow \angle PRQ = \angle PSQ
Thus , in both the cases, we have
\angle PRQ = \angle PSQ

(20) Prove that The angle in a semi-circle is a right angle.

Given: PQ is a diameter of a circle C(O,r) and \angle PRQ is an angle in semi-circle.
To Prove: \angle POQ = {90^ \circ }Proof: we know that the angle subtended by an arc of a circle at its centre is twice the angle formed by the same arc at a point on the circle. So, we have
\angle POQ = \angle PRQ
 \Rightarrow {180^ \circ } = 2\angle PRQ                                    [POQ is a straight line]
 \Rightarrow \angle PRQ = {90^ \circ }

(21) Prove that The opposite angles of a cyclic quadrilateral are supplementary.

Given: A Cyclic quadrilateral ABCD
To Prove: \angle A + \angle C = {180^ \circ } and \angle B + \angle D = {180^ \circ }
Construction: Join AC and BD.

Proof: Consider side AB of quadrilateral ABCD as the Chord of the circle. Clearly, \angle ACB and \angle ADB are angles in the same segment determined by chord AB of the Circle.
\angle ACB = \angle ADB          ………….(i)
Now , consider the side BC of quadrilateral ABCD as the chord of the circle. We find that \angle BAC  and \angle BDC are angles in the same segment
\angle BAC  = \angle BDC             [angles in the same segment are equal]..(ii)
Adding equation (i) and (ii), we get
 \Rightarrow        \angle ACB + \angle BAC = \angle ADB + \angle BDC
 \Rightarrow        \angle ACB + \angle BAC = \angle ADC
 \Rightarrow        \angle ABC + \angle ACB + \angle BAC = \angle ABC + \angle ADC
 \Rightarrow        {180^ \circ } = \angle ABC + \angle ADC         [sum of angle of triangle is {180^ \circ } ]
 \Rightarrow        \angle ABC + \angle ADC = {180^ \circ }
 \Rightarrow        \angle B + \angle D = {180^ \circ }
But, \angle A + \angle B + \angle C + \angle D = {360^ \circ }
\angle A + \angle C = {360^ \circ } - (\angle B + \angle D)
 \Rightarrow \angle A + \angle C = {360^ \circ } - {180^ \circ } = {180^ \circ }
Hence, \angle A + \angle C = {180^ \circ } and \angle B + \angle D= {180^ \circ }
The converse of this theorem is also true as given below.

(22) Prove that If the sum of any pair of opposite angles of a quadrilateral is {180^ \circ },then it is cyclic.

Given: A quadrilateral ABCD in which \angle B + \angle D = {180^ \circ }
To Prove: ABCD is acyclic quadrilateral.Proof: If possible, Let ABCD be not cyclic quadrilateral. Draw a circle passing through three non-collinear points A, B and C. Suppose the circle meets AD or AD produced at {D^{'}}. Join {D^{'}}C.
Now, ABCD’ is a cyclic quadrilateral.
\angle ABC + \angle A{D^{'}}C = {180^ \circ }..............(i)
But, \angle B + \angle D= {180^ \circ }
i.e. \angle ABC + \angle ADC = {180^ \circ }..............(ii)
from (i) and (ii), we get
\angle ABC + \angle A{D^{'}}C = \angle ABC + \angle ADC
 \Rightarrow        \angle A{D^{'}}C = \angle ADC
 \Rightarrow        An exterior angle of \Delta CD{D^{'}} is equal to interior oppsite angle.
But, this is not possible, unless {D^{'}} coincides with D. Thus, the circle passing through A,B,C also passes through D.
Hence, ABCD is a cyclic Quadrilateral.

(23) Prove that If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

Given: A Cyclic quadrilateral ABCD one of whose side AB is produced to E.
To Prove: \angle CBE = \angle ADCProof: Since ABCD is a quadrilateral and the sum of opposite pairs of angles in a cyclic quadrilateral is {180^ \circ }
\angle ABC + \angle ADC = {180^ \circ }
But,     \angle ABC + \angle CBE= {180^ \circ }                                  [Liner Pairs]
\angle ABC + \angle ADC = \angle ABC + \angle CBE
 \Rightarrow        \angle ADC = \angle CBE
Or,                               \angle CBE = \angle ADC

(24) Prove that The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

Given: A Cyclic quadrilateral ABCD in which AP,BP,CR and DR are the bisectors of \angle A, \angle B, \angle C and \angle D respectively such that a quadrilateral PQRS is formed.
To Prove: PQRS is a cyclic quadrilateral.Proof: In order to prove that PQRS is a cyclic quadrilateral, it is sufficinet to show that
\angle APB + \angle CRD = {180^ \circ }
Since the sum of the angles of a triangle is {180^ \circ }. Therefore, in triangles APB and CRD, we have
\angle APB + \angle PAB + \angle PBA = {180^ \circ }
And,                \angle CRD + \angle RCD + \angle RDC = {180^ \circ }
 \Rightarrow        \angle APB + {1 \over 2}\angle A + {1 \over 2}\angle B = {180^ \circ }
And,                \angle CRD + {1 \over 2}\angle C + {1 \over 2}\angle D = {180^ \circ }
 \Rightarrow        \angle APB + {1 \over 2}\angle A + {1 \over 2}\angle B + \angle CRD + {1 \over 2}\angle C + {1 \over 2}\angle D = {180^ \circ } + {180^\circ }
\angle APB + \angle CRD + {1 \over 2}\left\{ {\angle A + \angle B + \angle C + \angle D } \right\} = {360^ \circ }
\angle APB + \angle CRD + {1 \over 2}\left\{ {(\angle A + \angle C) + (\angle B + \angle D)} \right\} = {360^ \circ }
\angle APB + \angle CRD + {1 \over 2}({180^\circ } + {180^\circ }) = {360^ \circ }
\angle APB + \angle CRD = {180^\circ }
Hence, PQRS is a cyclic Quadrilateral.

(25) Prove that If two sides cyclic quadrilateral are parallel, then the remaining two sides are equal and the diagonals are also equal.

Given: A Cyclic quadrilateral ABCD in which AB\parallel DC.
To Prove: (i) AD = BC          (ii) AC = BDProof: In order to prove the desired results, it is sufficient to show that \Delta ADC \cong \Delta BCD. Since ABCD is cyclic Quadrilateral and sum of oppsite pairs of angles in a cyclic Quadrilateral is {180^ \circ }
\angle B + \angle D = {180^\circ }........(i)
Since AB\parallel DC and BC is a transveral and sum of the interior angles on the same side of a transversal is {180^\circ }
\angle ABC + \angle BCD = {180^\circ }
\angle B + \angle C = {180^\circ }...................(ii)
From (i) and (ii), we get
\angle B + \angle D = \angle B + \angle C
 \Rightarrow        \angle C = \angle D.................(iii)
Now, consider triangles ADC and BCD. In \Delta ADC and \Delta BCD, we have
\angle ADC= \angle BCD           [From equation (iii)]
DC=DC                                  [Common]
And,                \angle DAC= \angle CBD           [\angle DAC and \angle CBD are angles in the segment of chord CD]
So, by AAS-criterion of congruence, we have
\Delta ADC \cong \Delta BCD
 \Rightarrow       AD = BC and AC = BD

(26) Prove that If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.

Given: A cyclic quadrilateral ABCD such that AD = BC.
To Prove: AB\parallel CD
Construction: Join BD.Proof: We have,
AD = BC
 \Rightarrow        \mathop {DA}\limits^\frown \cong \mathop {BC}\limits^\frown
 \Rightarrow        \mathop {m(DA)}\limits^\frown \cong \mathop {(BC)}\limits^\frown
 \Rightarrow        2\angle 2 = 2\angle 1
 \Rightarrow        \angle 2 = \angle 1
But, these are alternate interior angles. Therefore, AB\parallel CD.

(27) Prove that An isosceles trapezium is cyclic.

Given: A trapezium ABCD in which AB\parallel DC and AD = BC
To Prove: ABCD is a cyclic trapezium.
Construction: Draw DE \bot AB and CF \bot AB.Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that \angle B + \angle D = {180^\circ }.
In triangles DEA and CFB, we have
AD = BC                        [Given]
\angle DEA = \angle CFB           [Each equal to {90^ \circ }]
And,                DE= CF
So, by RHS-criterion of congruence, we have
\Delta DEA \cong \Delta CFB
 \Rightarrow        \angle A = \angle B and \angle ADE = \angle BCF
Now,               \angle ADE = \angle BCF
 \Rightarrow        {90^ \circ } + \angle ADE = {90^ \circ } + \angle BCF
 \Rightarrow        \angle EDC + \angle ADE = \angle FCD + \angle BCF
 \Rightarrow        \angle ADC = \angle BCD
 \Rightarrow        \angle D = \angle C
Thus,               \angle A = \angle B and \angle C = \angle D.
\angle A + \angle B + \angle C + \angle D = {360^\circ }
 \Rightarrow        2\angle B + 2\angle D = {360^\circ }
 \Rightarrow        \angle B + \angle D = {180^\circ }
Hence, ABCD is a cyclic quadrilateral.



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