Exercise 10.1 : Solutions of Questions on Page Number : 171
Q1 :
Fill in the blanks
(i) The centre of a circle lies in __________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a __________ of the circle.
(iv) An arc is a __________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and __________ of the circle.
(vi) A circle divides the plane, on which it lies, in __________ parts.
Answer :
(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semi-circle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.
Q2 : Write True or False: Give reasons for your answers.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Answer :
(i) True. All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle.
(ii) False. There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.
(iii) False. Consider three arcs of same length as AB, BC, and CA. It can be observed that for minor arc BDC, CAB is a major arc. Therefore, AB, BC, and CA are minor arcs of the circle.
(iv) True. Let AB be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle. Therefore, it will be the diameter of the circle.
(v) False. Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example , in the given figure ,OAB is the sector of the circle
Exercise 10.2 : Solutions of Questions on Page Number : 173
Q1 : Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Answer :
A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius.
Consider two congruent circles having centre O and O’ and two chords AB and CD of equal lengths.
In ΔAOB and ΔCO’D,
AB = CD (Chords of same length)
OA = O’C (Radii of congruent circles)
OB = O’D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO’D (SSS congruence rule)
⇒ ∠ AOB = ∠ CO’D (By CPCT)
Hence, equal chords of congruent circles subtend equal angles at their centres.
Q2 : Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Answer :
Let us consider two congruent circles (circles of same radius) with centres as O and O’.
In ΔAOB and ΔCO’D,
∠ AOB = ∠ CO’D (Given)
OA = O’C (Radii of congruent circles)
OB = O’D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO’D (SAS congruence rule)
⇒ AB = CD (By CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Exercise 10.3 : Solutions of Questions on Page Number : 176
Q1 : Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Answer :
Consider the following pair of circles.
The above circles do not intersect each other at any point. Therefore, they do not have any point in common.
The above circles touch each other only at one point Y. Therefore, there is 1 point in common.
The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.
These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.
Q2 : Suppose you are given a circle. Give a construction to find its centre.
Answer :
The below given steps will be followed to find the centre of the given circle.
Step1. Take the given circle.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords.
Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.
Q3 : If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.
Answer :
Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.
Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O.
Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will also pass through O’.
Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.
Exercise 10.4 : Solutions of Questions on Page Number : 179
Q1 : Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Answer :
Let the radius of the circle centered at O and O’ be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O’A = O’B = 3 cm
OO’ will be the perpendicular bisector of chord AB.
∴ AC = CB
It is given that, OO’ = 4 cm
Let OC be x. Therefore, O’C will be x – 4
In ΔOAC,
OA2 = AC2 + OC2
⇒ 52 = AC2 + x2
⇒ 25 – x2 = AC2 … (1)
In ΔO’AC,
O’A2 = AC2 + O’C2
⇒ 32 = AC2 + (x – 4)2
⇒ 9 = AC2 + x2 + 16 – 8x
⇒ AC2 = – x2 – 7 + 8x … (2)
From equations (1) and (2), we obtain
25 – x2 = – x2 – 7 + 8x
8x = 32
x = 4
Therefore, the common chord will pass through the centre of the smaller circle i.e., O’ and hence, it will be the diameter of the smaller circle.
AC2 = 25 – x2 = 25 – 42 = 25 – 16 = 9
∴ AC = 3 cm
Length of the common chord AB = 2 AC = (2 x 3) cm = 6 cm
Q2 : If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer :
Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ VT = UT (By CPCT) … (1)
It is given that,
PQ = RS … (2)
⇒
⇒ PV = RU … (3)
On adding equations (1) and (3), we obtain
PV + VT = RU + UT
⇒ PT = RT … (4)
On subtracting equation (4) from equation (2), we obtain
PQ – PT = RS – RT
⇒ QT = ST … (5)
Equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.
Q3 : If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer :
Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠ OVT = ∠ OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ ∠ OTV = ∠ OTU (By CPCT)
Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.
Q4 : If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure 10.25).
Answer :
Let us draw a perpendicular OM on line AD.
It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
∴ BM = MC … (1)
And, AM = MD … (2)
On subtracting equation (2) from (1), we obtain
AM – BM = MD – MC
⇒ AB = CD
Q5 : Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Answer :
Draw perpendiculars OA and OB on RS and SM respectively.
OR = OS = OM = 5 m. (Radii of the circle)
In ΔOAR,
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 – 9) m2 = 16 m2
OA = 4 m
ORSM will be a kite (OR = OM and RS = SM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
∴∠RCS will be of 90° and RC = CM
Area of ΔORS =
Therefore, the distance between Reshma and Mandip is 9.6 m.
Q6 : A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Answer :
It is given that AS = SD = DA
Therefore, ΔASD is an equilateral triangle.
OA (radius) = 20 m
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write
∴ AB = OA + OB = (20 + 10) m = 30 m
In ΔABD,
AD2 = AB2 + BD2
AD2 = (30)2 +
Therefore, the length of the string of each phone will be m
.
Exercise 10.5 : Solutions of Questions on Page Number : 184
Q1 : In the given figure, A, B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.
Answer :
It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
Q2 : A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Answer :
In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60°
In cyclic quadrilateral ACBD,
∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)
⇒ ∠ADB = 180° – 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.
Q3 : In the given figure, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.
Answer :
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
PQRS is a cyclic quadrilateral.
∠ PQR + ∠ PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠ PSR = 180° – 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠ POR = 2∠ PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠ OPR = ∠ ORP (Angles opposite to equal sides of a triangle)
∠ OPR + ∠ ORP + ∠ POR = 180° (Angle sum property of a triangle)
2 ∠ OPR + 160° = 180°
2 ∠ OPR = 180° – 160° = 20º
∠ OPR = 10°
Q4 : In the given figure, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
Answer :
In ΔABC,
∠ BAC + ∠ ABC + ∠ ACB = 180° (Angle sum property of a triangle)
⇒ ∠ BAC + 69° + 31° = 180°
⇒ ∠ BAC = 180° – 100º
⇒ ∠ BAC = 80°
∠ BDC = ∠ BAC = 80° (Angles in the same segment of a circle are equal)
Q5 : In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.
Answer :
In ΔCDE,
∠ CDE + ∠ DCE = ∠ CEB (Exterior angle)
⇒ ∠ CDE + 20° = 130°
⇒ ∠ CDE = 110°
However, ∠ BAC = ∠ CDE (Angles in the same segment of a circle)
⇒ ∠ BAC = 110°
Q6 : ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.
Answer :
For chord CD,
∠ CBD = ∠ CAD (Angles in the same segment)
∠ CAD = 70°
∠ BAD = ∠ BAC + ∠ CAD = 30° + 70° = 100°
∠ BCD + ∠ BAD = 180° (Opposite angles of a cyclic quadrilateral)
∠ BCD + 100° = 180°
∠ BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠ BCA = ∠ CAB (Angles opposite to equal sides of a triangle)
⇒ ∠ BCA = 30°
We have, ∠ BCD = 80°
⇒ ∠ BCA + ∠ ACD = 80°
30° + ∠ ACD = 80°
⇒ ∠ ACD = 50°
⇒ ∠ ECD = 50°
Q7 : If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Answer :
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as a chord)
∠BCD + ∠BAD = 180° (Cyclic quadrilateral)
∠BCD = 180° – 90° = 90°
(Considering AC as a chord)
∠ADC + ∠ABC = 180° (Cyclic quadrilateral)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.
Q8 : If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer :
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
AD = BC (Given)
∠ AMD = ∠ BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ ADC = ∠ BCD (CPCT) … (1)
∠ BAD and ∠ ADC are on the same side of transversal AD.
∠ BAD + ∠ ADC = 180° … (2)
∠ BAD + ∠ BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
Q9 : Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ ACP = ∠ QCD.
Answer :
Join chords AP and DQ.
For chord AP,
∠ PBA = ∠ ACP (Angles in the same segment) … (1)
For chord DQ,
∠ DBQ = ∠ QCD (Angles in the same segment) … (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠ PBA = ∠ DBQ (Vertically opposite angles) … (3)
From equations (1), (2), and (3), we obtain
∠ ACP = ∠ QCD
Q10 : If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Answer :
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
Join AD.
∠ ADB = 90° (Angle subtended by semi-circle)
∠ ADC = 90° (Angle subtended by semi-circle)
∠ BDC = ∠ ADB + ∠ ADC = 90° + 90° = 180°
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC.
Q11 : ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.
Answer :
In ΔABC,
∠ ABC + ∠ BCA + ∠ CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ BCA + ∠ CAB = 180°
⇒ ∠ BCA + ∠ CAB = 90° … (1)
In ΔADC,
∠ CDA + ∠ ACD + ∠ DAC = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ ACD + ∠ DAC = 180°
⇒ ∠ ACD + ∠ DAC = 90° … (2)
Adding equations (1) and (2), we obtain
∠ BCA + ∠ CAB + ∠ ACD + ∠ DAC = 180°
⇒ (∠ BCA + ∠ ACD) + (∠ CAB + ∠ DAC) = 180°
∠ BCD + ∠ DAB = 180° … (3)
However, it is given that
∠ B + ∠ D = 90° + 90° = 180° … (4)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠ CAD = ∠ CBD (Angles in the same segment)
Q12 : Prove that a cyclic parallelogram is a rectangle.
Answer :
Let ABCD be a cyclic parallelogram.
∠ A + ∠ C = 180° (Opposite angles of a cyclic quadrilateral) … (1)
We know that opposite angles of a parallelogram are equal.
∴ ∠ A = ∠ C and ∠ B = ∠ D
From equation (1),
∠ A + ∠ C = 180°
⇒ ∠ A + ∠ A = 180°
⇒ 2 ∠ A = 180°
⇒ ∠ A = 90°
Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.
Exercise 10.6 : Solutions of Questions on Page Number : 186
Q1 : Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer :
Let two circles having their centres as O and intersect each other at point A and B respectively. Let us join O.
In ΔAO and BO,
OA = OB (Radius of circle 1)
A = B (Radius of circle 2)
O = O (Common)
ΔAO ≅ ΔBO (By SSS congruence rule)
∠OA = ∠OB (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Q2 : Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer :
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
(Perpendicular from the centre bisects the chord)
Let ON be x. Therefore, OM will be 6 – x.
In ΔMOB,
In ΔNOD,
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
From equation (2),
Therefore, the radius of the circle is
cm.
Q3 : The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Answer :
Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB =
In ΔOMB,
In ΔOND,
Therefore, the distance of the bigger chord from the centre is 3 cm.
Q4 : Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer :
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
AD = CE (Given)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) … (1)
∠ODA = ∠OEC (By CPCT) … (2)
Also,
∠OAD = ∠ODA (As OA = OD) … (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180 º – 2x – a … (4)
However, ∠DOE = 180 º – 2y
And, ∠AOC = 180 º – 2a
∠DOE – ∠AOC = 2a – 2y = 2a – 2 (180 º – 2x – a)
= 4a + 4x – 360° … (5)
∠BAC + ∠CAD = 180 º (Linear pair)
⇒ ∠BAC = 180 º – ∠CAD = 180 º – (a + x)
Similarly, ∠ACB = 180 º – (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180 º (Angle sum property of a triangle)
∠ABC = 180 º – ∠BAC – ∠ACB
= 180 º – (180 º – a – x) – (180 º – a – x)
= 2a + 2x – 180 º
=
[4a + 4x – 360°]
∠ABC = [∠DOE – ∠ AOC] [Using equation (5)]
Q5 : Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer :
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠ COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠ AOB = ∠ BOC = ∠ COD = ∠ DOA = 90°
Clearly, point O has to lie on the circle.
Q6 : ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Answer :
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠ AEC + ∠ CBA = 180°
∠ AEC + ∠ AED = 180° (Linear pair)
∠ AED = ∠ CBA … (1)
For a parallelogram, opposite angles are equal.
∠ ADE = ∠ CBA … (2)
From (1) and (2),
∠ AED = ∠ ADE
AD = AE (Angles opposite to equal sides of a triangle)
Q7 : AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Answer :
Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
OA = OC (Given)
OB = OD (Given)
∠ AOB = ∠ COD (Vertically opposite angles)
ΔAOB ≅ ΔCOD (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that ΔAOD ≅ ΔCOB
∴ AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
∴ ∠ A = ∠ C
However, ∠ A + ∠ C = 180° (ABCD is a cyclic quadrilateral)
⇒ ∠ A + ∠ A = 180°
⇒ 2 ∠ A = 180°
⇒ ∠ A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.
∠ A is the angle subtended by chord BD. And as ∠ A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
Q8 : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°
.
Answer :
It is given that BE is the bisector of ∠B.
∴ ∠ABE =
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
⇒ ∠ADE =
Similarly, ∠ACF = ∠ADF =
(Angle in the same segment for chord AF)
∠D = ∠ADE + ∠ADF
Similarly, it can be proved that
Q9 : Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Answer :
AB is the common chord in both the congruent circles.
∴ ∠ APB = ∠ AQB
In ΔBPQ,
∠ APB = ∠ AQB
∴ BQ = BP (Angles opposite to equal sides of a triangle)
Q10 : In any triangle ABC, if the angle bisector of ∠ A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer :
Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.
Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 ∠BAC = 2 ∠A … (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHS congruence rule)
∠BOE = ∠COE (By CPCT) … (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A … (3)
Since AD is the bisector of angle ∠A,
∠BAD =
⇒ 2 ∠BAD = ∠A … (4)
From equations (3) and (4), we obtain
∠BOD = 2 ∠BAD
This can be possible only when point BD will be a chord of the circle. For this, the point D lies on the circum circle.
Therefore, the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circum circle of triangle ABC.