# 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 $OQ$ Proof: 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 $OQ$ Proof: 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 ${O^{'}}B$ Proof: 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 $AB = CD$and $OL \bot AB$ 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 OC Proof: 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: $\angle AOB = \angle COD$ Proof: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: $AB = CD$ Proof: 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 $OQ$ Proof: 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.

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: $\angle CBE = \angle ADC$ Proof: 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) $AC = BD$ Proof: 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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