Answer: AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...

### In the given, AB is the diameter of the circle with centre O. If ∠ADC = 32o, find angle BOC.

Solution: \[Arc\text{ }AC\text{ }subtends\angle AOC\]at the centre and \[\angle ADC\]at the remaining part of the circle. Thus, \[\angle AOC\text{ }=\text{ }2\angle ADC\] \[\angle AOC\text{ }=\text{...

### In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Solution: Join \[EB\] Then, in cyclic quad. \[ABEP\] \[\angle APE\text{ }+\angle ABE\text{ }=\text{ }{{180}^{o}}~\ldots ..\text{ }\left( i \right)\][Opposite angles of a cyclic quad. are...

### In the given figure, AB = AD = DC = PB and ∠DBC = xo. Determine, in terms of x: (i) ∠ABD, (ii) ∠APB. Hence or otherwise, prove that AP is parallel to DB.

Solution: Given, \[AB\text{ }=\text{ }AD\text{ }=\text{ }DC\text{ }=\text{ }PB\text{ }and~\angle DBC\text{ }=\text{ }{{x}^{o}}\] Join \[AC\text{ }and\text{ }BD\] Proof: \[\angle DAC\text{ }=\angle...

### If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66o and ∠ABC = 80o. Calculate: ∠BIC

Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...

### If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66o and ∠ABC = 80o. Calculate: (i) ∠DBC, (ii) ∠IBC,

Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...

### In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92o, ∠FAE = 20o; determine ∠BCD. Given reason in support of your answer.

Solution: Given, In cyclic quad. \[ABCD\] \[AF\text{ }||\text{ }CB\text{ }and\text{ }DA\] is produced to \[E\]such that \[\angle ADC\text{ }=\text{ }{{92}^{o}}~and\angle FAE\text{ }=\text{...

### D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.

Given, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\text{ }and\text{ }D\text{ }and\text{ }E\]are points on \[AB\text{ }and\text{ }AC\]such that \[AD\text{ }=\text{ }AE\] And, \[DE\] is joined....

### In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate: ∠BAC

Solution: In cyclic quadrilateral \[ABEC\] \[\angle BAC\text{ }+\angle BEC\text{ }=\text{ }{{180}^{o}}~\] [Opposite angles of a cyclic quadrilateral are supplementary] \[\angle BAC\text{ }+\text{...

### In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate: (i) ∠BDC (ii) ∠BEC

Solution: (i) Given that \[BD\] is a diameter of the circle. And, the angle in a semicircle is a right angle. So, \[\angle BCD\text{ }=\text{ }90{}^\circ \] Also given that, \[\angle DBC\text{...

### The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Let \[ABCD\] be a cyclic quadrilateral and \[PQRS\] be the quadrilateral formed by the angle bisectors of angle \[\angle A,\angle B,\angle C\text{ }and\angle D\] Required to prove: \[PQRS\] is a...

### In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65o, calculate ∠DEC.

Solution: Join \[OE\] \[Arc\text{ }EC\] subtends \[\angle EOC\]at the centre and \[\angle EBC\]at the remaining part of the circle. \[\angle EOC\text{ }=\text{ }2\angle EBC\text{ }=\text{ }2\text{...

### Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Suppose, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\]and circle with \[AB\] as diameter is drawn which intersects the side \[BC\text{ }and\text{ }D\] And, join \[AD\] Proof: It’s seen that,...

### In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.

Solution: Firstly, join \[OB\text{ }and\text{ }OC\] Proof: \[\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\] Now, in \[\vartriangle OBC\]...

### In the given circle with diameter AB, find the value of x.

Solution: Now, \[\angle ABD\text{ }=\angle ACD\text{ }=\text{ }{{30}^{o}}~\][Angles in the same segment] In \[\vartriangle ADB\] by angle sum property we have \[\angle BAD\text{ }+\angle ADB\text{...

### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: ∠ABC.

Solution: According to the given question, \[AC\] is the side of a regular octagon, \[\angle AOC\text{ }=\text{ }{{360}^{o}}/\text{ }8\text{ }=\text{ }{{45}^{o}}\] Hence, \[arc\text{ }AC\]subtends...

### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: (i) ∠AOB, (ii) ∠ACB,

Solution: (i) \[Arc\text{ }AB\] subtends\[\angle AOB\]at the centre and \[\angle ACB\]at the remaining part of the circle. \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB\]...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: ∠PQR.

Join \[OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS\] Given, \[PQ\text{ }=\text{ }QR\text{ }=\text{ }RS\] So, \[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\] [Equal chords subtends equal...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: (i) ∠POS, (ii) ∠QOR,

Solution: Join \[OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS\] Given, \[PQ\text{ }=\text{ }QR\text{ }=\text{ }RS\] So, \[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\] [Equal chords subtends...

### If two sides of a cycli-quadrilateral are parallel; prove that: (i) its other two sides are equal. (ii) its diagonals are equal.

Let ABCD is a cyclic quadrilateral in which\[AB\text{ }||\text{ }DC\] \[AC\text{ }and\text{ }BD\]are its diagonals. Required to prove: \[\left( i \right)\text{ }AD\text{ }=\text{ }BC\] \[\left( ii...

### In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110o, calculate: (i) ∠AFE, (ii) ∠FAB.

Solution: Join \[AE,\text{ }OB\text{ }and\text{ }OC\] (i) As \[AOD\]is the diameter \[\angle AED\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] But, given \[\angle DEF\text{...

### In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

Solution: Let \[ABCD\]be the cyclic trapezium in which \[AB\text{ }||\text{ }DC,\text{ }AC\text{ }and\text{ }BD\]are the diagonals. Required to prove: \[\left( i \right)\text{ }AD\text{ }=\text{...

### ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

We assume that \[ABCD\]be the given cyclic quadrilateral. \[PA\text{ }=\text{ }PD\text{ }\left[ Given \right]\] So, \[\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)\]...

### Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution: Join \[AC,\text{ }PQ\text{ }and\text{ }BD.\] As \[ACQP\]is a cyclic quadrilateral \[\angle CAP\text{ }+\angle PQC\text{ }=\text{ }{{180}^{o}}~\ldots \ldots .\text{ }\left( i \right)\]...

### In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25o; find the angle AEB. Give reasons in support of your answer.

Solution: Join \[AC\text{ }and\text{ }BD.\] Hence, we get \[\angle CAD\text{ }=\text{ }{{90}^{o}}~and\angle CBD\text{ }=\text{ }{{90}^{o}}\] [Angle is a semicircle is a right angle] And, \[AB\text{...

### In the figure given RS is a diameter of the circle. NM is parallel to RS and angle MRS = 29degree Calculate: (i) ∠RNM; (ii) ∠NRM.

Solution: (i) Join \[RN\text{ }and\text{ }MS\] \[\angle RMS\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] By angle sum property of \[\vartriangle RMS\] \[\angle RMS\text{...

### Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the center O is equal to twice the angle APC

Solution: To prove: \[\angle AOC\text{ }+\angle BOD\text{ }=\text{ }2\angle APC\] \[OA,\text{ }OB,\text{ }OC\text{ }and\text{ }OD\] are joined. \[AD\] is joined. Now, it’s seen that \[\angle...

### The figure given below, shows a circle with centre O. Given: ∠AOC = a and ∠ABC = b. (i) Find the relationship between a and b (ii) Find the measure of angle OAB, if OABC is a parallelogram.

Solution: (i) According to the given question, it’s clear that \[\angle ABC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Reflex\text{ }(\angle COA)\] [Angle at the centre is double the...

### Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

Solution: According to the given question, Let \[O\text{ }and\text{ }O\]be the centres of two intersecting circles, where points of the intersection are \[P\text{ }and\text{ }Q\text{ }and\text{...

### In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.

Solution: According to the given question, \[AB\text{ }=\text{ }AC\] So, \[\angle B\text{ }=\angle C\text{ }\ldots \text{ }\left( 1 \right)\] [Angles opposite to equal sides are equal] And,...

### Prove that: (i) the parallelogram, inscribed in a circle, is a rectangle. (ii) the rhombus, inscribed in a circle, is a square.

Solution: (i) Let’s expect that \[ABCD\]is a parallelogram which is inscribed in a circle. Hence, we know that \[\angle BAD\text{ }=\angle BCD\] [Opposite angles of a parallelogram are equal] And...

### ABCD is a parallelogram. A circle

Solution: Now, \[\angle BAD\text{ }+\angle BFE\text{ }=\text{ }{{96}^{o}}~+\text{ }{{84}^{o}}~=\text{ }{{180}^{o}}\] But these two are interior angles on the same side of a pair of lines \[AD\text{...

### In the following figure, (i) if ∠BAD = 96degree, find ∠BCD and ∠BFE. (ii) Prove that AD is parallel to FE.

Solution: \[ABCD\]is a cyclic quadrilateral (i) Hence, \[\angle BAD\text{ }+\angle BCD\text{ }=\text{ }{{180}^{o}}\] [Pair of opposite angles in a cyclic quadrilateral are supplementary] \[\angle...

### In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.

Solution: Join \[OB.\] Then,\[\angle OBA\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle is a right angle] Which means, \[OB\]is perpendicular to \[AE.\] Now, we know that the perpendicular...

### In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80degree and ∠CDE = 40degree. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.

Solution: (i) According to the given question, We know that \[\angle DCE\text{ }=\text{ }{{90}^{o}}~\angle CDE\] \[=\text{ }{{90}^{o}}-\text{ }{{40}^{o}}~=\text{ }{{50}^{o}}\] Hence, \[\angle...

### In ABCD is a cyclic quadrilateral in which ∠DAC = 27degree, ∠DBA = 50degree and ∠ADB = 33degree. Calculate ∠CAB.

Solution: In quad. \[ABCD,\] \[\angle DAB\text{ }+\angle DCB\text{ }=\text{ }{{180}^{o}}\] \[{{27}^{o}}~+\angle CAB\text{ }+\text{ }{{83}^{o}}~=\text{ }{{180}^{o}}\] Hence, \[\angle CAB\text{...

### In ABCD is a cyclic quadrilateral in which ∠DAC = 27degree, ∠DBA = 50degree and ∠ADB = 33degree. Calculate (i) ∠DBC, (ii) ∠DCB

Solution: (i) According to the given question, We know that \[\angle DBC\text{ }=\angle DAC\text{ }=\text{ }{{27}^{o}}\] [Angles subtended by the same chord on the circle are equal] (ii)...

### In the following figure, O is the centre of the circle; ∠AOB = 60degree and ∠BDC = 100degree, find ∠OBC.

Solution: According to the given question, we get \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB\] \[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{...

### In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110degree, find ∠BDC.

Solution: Join \[AD\] So, we get \[\angle ADC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{110}^{o}}~=\text{...

### ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130degree, find ∠BAC.

Solution: According to the given question \[\angle ACB\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is 90o] Also, \[\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle ADC\] \[=\text{...

### Given: ∠CAB = 75degree and ∠CBA = 50degree. Find the value of ∠DAB + ∠ABD.

Solution: According to the given question, \[\angle CAB\text{ }=\text{ }{{75}^{o}}~and\angle CBA\text{ }=\text{ }{{50}^{o}}\] In \[\vartriangle ABC\], by angle sum property we have \[\angle...

### In the figure given below, O is the centre of the circle and triangle ABC is equilateral. Find: (i) ∠ADB, (ii) ∠AEB

Solution: (i) According to the given question, \[\angle ACB\text{ }and\angle ADB\]are in the same segment, So, \[\angle ADB\text{ }=\angle ACB\text{ }=\text{ }{{60}^{o}}\] (ii) Now, join \[OA\text{...

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75degree; ∠ABD = 58degree and ∠ADC = 77degree. Find ∠BCA.

Solution: According to the given ques, \[\angle BCA\text{ }=\angle ADB\text{ }=\text{ }{{47}^{o}}\] [Angles subtended by the same chord on the circle are equal]

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75Degree; ∠ABD = 58degree and ∠ADC = 77degree. Find: (i) ∠BDC, (ii) ∠BCD,

Solution: (i) According to the given question, By angle sum property of triangle ABD, \[\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{75}^{o}}-\text{ }{{58}^{o}}~=\text{ }{{47}^{o}}\] Hence,...

### Calculate: ∠ACB.

Solution: According to the given question By angle sum property of a triangle we have \[\angle ACB\text{ }=\text{ }{{180}^{o}}-\text{ }{{49}^{o}}-\text{ }{{43}^{o}}~=\text{ }{{88}^{o}}\]

### Calculate: (i) ∠CDB, (ii) ∠ABC,

Solution: According to the given question We get, (i) \[\angle CDB\text{ }=\angle BAC\text{ }=\text{ }{{49}^{o}}\] (ii) \[\angle ABC\text{ }=\angle ADC\text{ }=\text{ }{{43}^{o}}\] [Angles subtended...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

Solution: According to the given question, (i) In \[\vartriangle AOB,\]we have \[OA\text{ }=\text{ }OB\text{ }\left( radii \right)\] Thus, \[\angle OBA\text{ }=\angle OAB\] By angle sum property of...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠ACB, (ii) ∠OBC,

Solution: According to the given question, Given, \[\angle AOB\text{ }=\text{ }{{140}^{o}}~\]and \[\angle OAC\text{ }=\text{ }{{50}^{o}}\] (i) So, \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle...

### In the figure, given below, find: ∠ABC. Show steps of your working.

Solution: The sum of angles in a quadrilateral is \[{{360}^{o}}\] Thus, \[\angle ADC\text{ }+\angle DAB\text{ }+\angle BCD\text{ }+\angle ABC\text{ }=\text{ }{{360}^{o}}\] \[{{75}^{o~}}+\text{...

### In the figure, given below, find: (i) ∠BCD, (ii) ∠ADC, Show steps of your working.

Solution: According to the given question, it’s clear that In cyclic quadrilateral \[ABCD,\text{ }DC\text{ }||\text{ }AB\] And given, \[\angle DAB\text{ }=\text{ }{{105}^{o}}\] (i) Hence, \[\angle...

### In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centers of two circles.

Solution: According to the given question, it’s clear that \[\angle DBA\text{ }=\angle CBA\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] So, adding both \[\angle DBA\text{...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

(i) (ii) SOLUTION: (i) According to the given question, \[\angle AOB\text{ }=\text{ }2\angle AOB\text{ }=\text{ }2\text{ }x\text{ }{{50}^{o}}~=\text{ }{{100}^{o}}\] [Angle at the center is double...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

SOLUTION: (i) In the following figure, \[\angle BAD\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle] Therefore, \[\angle BDA\text{ }=\text{ }{{90}^{o}}-\text{ }{{35}^{o}}~=\text{ }{{55}^{o}}\]...

### In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

(i) (ii) Solution: (i)In the following figure, \[b\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{130}^{o}}\] [Angle at the center is double the angle at the circumference...

### Given O is the centre of the circle and ∠AOB = 70o. Calculate the value of: (i) ∠OCA, (ii) ∠OAC.

Solution: \[\angle AOB\text{ }=\text{ }2\angle ACB\] [Angle at the center is double the angle at the circumference subtend by the same chord] \[\angle ACB\text{ }=\text{ }{{70}^{o}}/\text{ }2\text{...

### In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB.

Solution: (i) In \[\vartriangle ABD,\] \[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\] \[{{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]...

### In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30Degree and 40Degree respectively. Find ∠AOC Show your steps of working.

Solution: Let’s join \[AC.\] And, let \[\angle OAC\text{ }=\angle OCA\text{ }=\text{ }x\] [Angles opposite to equal sides are equal] Thus, \[\angle AOC\text{ }=\text{ }{{180}^{o}}-\text{ }2x\] Also,...