Book Name | : Ganit Prakash |
Subject | : Mathematics |
Class | : 10 (Madhyamik) |
Publisher | : Prof Nabanita Chatterjee |
Chapter Name | : Theorems Related To Cyclic Quadrilateral (10th Chapter) |
Application 1.
Look at the picture of two circles with centre O given below and 1 us write by calculating the value of x° in each case.
Solution:
(i) ABCD is a cyclic quadrilateral. \quad\therefore ∠ABC + ∠ADC 180° – 130° = 50°
\therefore x = 50.
(ii) ABCD is a cyclic quadrilateral. \quad\therefore ∠ABC + ∠ADC = 180°
\therefore ∠ABC + ∠ADC = 180°
\therefore∠ABC = 50°
Again, ∠BAC = one right angle
\therefore ∠BAC = 90°
\therefore x° + ∠ABC + ∠BAC = 180°
\therefore x° + 50° + 90° = 180° ; \quad \therefore x = 40°
Application 2.
ABCD is a cyclic quadrilateral and O is the centre of that circle. ∠COD = 120° and ∠BAC = 30°, let us write by caluclating the value of ∠BOC And ∠BCD.
Solution:
∠BOC is angle at the centre and ∠BAC is angle at the circle standing on the same minor arc BC.
\therefore∠BOC = 2 ∠BAC = 60°
∠COD is angle at the centre and ∠CAD is the angle at the circle standing on the same minor arc CD.
\therefore ∠CAD = \frac{1}{2} ∠COD = \frac{1}{2} × 120° = 60°
\therefore ∠BAC = 30° + 60° = 90°
\therefore In concyclic quadrilateral ABCD, ∠BCD + ∠BAD = 1
\therefore ∠BCD = 90°
Application 3.
Sides AD and AB of a cyclic quadrilateral ABCD beside are produced to the points E and F respectively. If ∠CBF = 120°, let us write by calculating the value of ∠CDE. [Do it yourself]
Solution:
In the figure, ∠BAD = ∠CBF = 120°
∠CDE = 180° – 120° = 60°
Application 4.
ABCD is a cyclic quadrilateral. Sides AB and DC, when produced, meet at the point Q. If ∠ADC = 85° and ∠BPC = 40°, let us write by calculating the value of ∠BAD and ∠COD.
Solution:
For cyclic quadrilateral ABCD, exterior ∠PBC = ∠ADC = 85°
\therefore In ∆BPC, ∠BCP = 180° – (85° + 40°) = 55°
Again, ∠BAD = exterior ∠BCP = 55°
In ∆CQD, ∠CQD + ∠DCQ = 85°
∠CQD = 85° – ∠DCQ = 85° – ∠BCP = 55°
∠BAD = 55° and ∠CQD = 30°
Application 5.
Let us prove that a cyclic parallelogram is a rectangular picture
Solution:
Given : ABCD is a cyclic parallelogram.
To prove : Quadrilateral ABCD is a rectangular picture.
Proof : ABCD is a parallelogram.
∠ABC = ∠ADC
Again, ABCD is a cyclic quadrilateral.
∠ABC + ∠ADC = 180°
So, 2 ∠ABC = 180° \therefore ∠ABC = 90°
One angle of the parallelogram is right angle. So, ABCD is a rectangular picture.
LET US SEE BY CALCULATING – 10
Question 1
In the picture beside two diagonals of quadrilateral PQRS intersect each other at the point X in such a way that ∠PRS = 65° and ∠RQS = 45°. Let us write by calculating value of ∠SQP and ∠RSP.
Solution:
∠SQP = ∠PRS = 65° (on the same segment) ∠RPS = ∠RQS = 40° (on the same segment)
∠RSP = 180° – (65° + 45°) = 180° – 110° = 70°
Question 2
Side AB of a cyclic quadrilateral ABCD is produced to the point X and by measuring we see that ∠XBC = 82° and ∠ADB = 47°. Let us write by measuring the value of ∠BAC.
Solution:
Find ∠BAC.
As ABCD is a cyclic quadrilateral.
∠ADC + ∠ABC = 2 rt. angles
Again, ∠ABC + ∠XBC = 2 rt. angle
\therefore ∠ADC + ∠ABC = ∠ABC + ∠XBC
∠ADC = ∠XBC = 80° (given)
\therefore ∠BDC + ∠BDA = ∠ADC
∠BDC + 47° = 80° [∠BDA = 47° (given)]
\therefore ∠BDC = 82° – 470° = 35°
∠BDC & ∠BACAre on the same segment.
∠BAC = ∠BDC = 35°
∠BAC = 35°
Question 3
Two sides PQ and SR of a cyclic quadrilateral PQRS are produced to meet at the oint T. O is the centre of the circle. If ∠POQ = 110°, ∠QOR = 60°, ∠ROS = 80°, let us Vrite by measuring ∠RQS and ∠OTR.
Solution:
PQ & RS, the two sides of the cyclic quadrilateral, meet at T, when produced. O is the centre of the circle.
∠POQ = 110°, ∠QOR = 60°, & R ∠OS = 80°
To find ∠RQS & ∠QTR.
On the same segment SR, ∠RQS is the angle on the circumference
∠ROS = \frac{1}{2} \text{ ∠ROS } = \frac{1}{2} × 80° = 40°
\therefore We know, ∠POQ + ∠QOR + ∠ROS + ∠SOP = 360°
\therefore ∠SOP = 360° – (∠POQ + ∠QOR + ∠ROS)
= 360° – (110° + 60° + 80°) = 360° – 250° = 110°
or, ∠SOP = 110°, ∠PQR = ∠PQS + ∠SQR = 55° + 40° = 95°
\therefore ∠PQR + ∠RQT = 180°
∠RQT = 180° – ∠PQR = 180° – 95° = 85°
In ∆Q OR, OQ = OR (same radiii)
\therefore ∠OQR = OR
Again, ∠OQR + ∠ORQ + ∠QOR = 180° or, 2 ∠OQR + 60° = 180°
∠OQR = \frac{1}{2}(180° – 60°) = 60° = ∠ORQ
Again, In ∆ORS, OR = OS (same radii)
\therefore ∠ORS = ∠OSR
∠ORS + ∠OSR + ∠SOR = 180°
2 ∠ORS + 80° = 180°
∠ORS = ∠OSR = \frac{1}{2}(180° – 80°) = 50°
∠SRQ = ∠ORS + ∠ORQ = 50° + 60° = 110°
∠QRT = 180° – & ∠SRQ = 180° – 110° = 70°
In ∆QTR, QTR + ∠QRT + ∠RQT = 180°
∠QTR = 180° – QRT + RQT
= 180° – (70° + 85°) = 25°
\therefore ∠RQS = 40° & ∠QTR = 25°
Question 4
Two circles intersect each other at the points P and Q. Two straight lines throug the points P and Q intersect one circle at points A and C respectively and the other circle at points B and O respectively. Let us prove that AC || BD.
Solution:
To prove AC || BD.
Join A, C ; B, D & P, Q.
Proof: ACQP is a cyclic quadrilateral.
\therefore ∠PAC + ∠PQC = 2 rt. angles
Again, PQ meet CD at Q
\therefore ∠PAC + ∠PQD = 2 rt. angles
\therefore ∠PAC + ∠PQC = ∠PQC + ∠PQD
∠PAC = ∠PQD
As PQDB is a cyclic quadrilateral.
\therefore ∠PBD + ∠PQD = 2 rt. angles
or, ∠PBD + ∠PAC = 2 rt. angles (\because ∠PQD = ∠PAC)
\therefore AB cuts AC & BD, and sum of the internal angles of the same sides of the intercel is equal to two right angle.
\therefore AC || BD
Question 5
I drew a cyclic quadrilateral AQCD and the side BC is produced to the point E. Let us prove that the bisectors of ∠BAD and ∠DCE meet in the circle.
Solution:
ABCD is a cyclic quadrilateral. Side BC is produced to E & the bisector ∠BAD cuts the circle at F.
Join C, F.
To prove, CF straight line bisects ∠DCE.
Proof : ∠DCF & ∠DAF are on the same segment.
\therefore ∠DCF = ∠DAF
AF bisects ∠BAD.
\therefore ∠DAF = \frac{1}{2} ∠BAD
\therefore ∠DCF = ∠DAF = \frac{1}{2} ∠BAD ——————– (i)
ABCD is a cyclic quadrilateral.
\therefore ∠BAD + ∠BCD = 2 rt angles ——————– (ii)
BE meets DCAt C.
∠DCE + ∠BCD = 2 rt angles ——————– (iii)
From (ii) & (iii),
∠BAD + ∠BCD = ∠DCE + ∠BCD
or, ∠BAD = ∠DCE ——————– (iv)
∠DCE = 2 ∠DCF
CF is the bisector of ∠DCE.
Question 6
Mohit drew two straight lines through any point X exterior to a circle to intersect the circle at points A, B and points C, D respectively. Let us prove logically that ∆XAC and ∆XBD are equiangular.
Solution:
From an external point X, two straight lines are drawn, which cut the circle at A, B and C, D. To prove that two angles of each of the ∆XAC & ∆XBD are equal, i.e., ∆XAC & ∆XBD are equiangular. Join A, C & B, D.
ABCD is a cyclic quadrilateral
\therefore ∠ABD + ∠ACD = 2 rt. angles ——————– (i)
AC meets DX at C
\therefore ∠ACD + ∠ACX = 2rt. angles ——————– (ii)
\therefore ∠ABD + ∠ACD = ∠ACD + ∠ACX
\therefore ∠ABD = ∠ACX = ∠XBC
Again, as ABCD is a cyclic quadrilateral,
∠BDC + ∠BAC = 2 rt. angles
& BAC + CAX = 2 rt. angles \therefore ∠BDC + ∠BAC = ∠BAC + ∠CAX
\therefore ∠BDC = ∠CAX or, ∠BDX = ∠CAX
Now, in ∆XAC & ∆XBD,
∠XCA = ∠XBD ; ∠XAC = ∠BDX & ∠AXC = ∠BXD (Same angle)
\therefore ∆XAC & ∆XBD are equiangular.
Question 7
I drew two circles which intersect each other at the points G and H. Now I drew a straight line through the point G which intersects two circles at the points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S. Let us prove that PQ = RS.
Solution:
Two circles intersect each other at G & H. A straight line passing through G cuts the circles at P & Q. And a straight line passing through H cuts the circles at R & S.
To prove PQ = RS.
Join P, R ; G, H & Q, S.
PRHG is a cyclic quadrilateral.
\therefore ∠RPG + ∠RHG = 2 rt. angles.
Again, GH meets RS at H.
\therefore ∠RHG + ∠GHS = 2 rt. angles
\therefore ∠RPG + ∠RHG = ∠RHG + ∠GHS
\therefore ∠RPG = ∠GHS ——————– (i)
Again, as GHSQ is a cyclic quadrilateral.
\therefore ∠GQS + ∠GHS = 2 it. angles. and ∠GHS + ∠GHR = 2 rt. angles.
\therefore ∠GQS + ∠GHS = ∠GHS + ∠GHR
∠GQS = ∠GHR ——————– (ii)
Adding (i) & (ii),
∠RPG + ∠GQS = ∠GHS + ∠GHR
or, ∠RPG + ∠GQS = 2 rt. angles
i.e., straight line PQ cuts PR & QS, & the sum of the internal angles on the same side the intercept are equal to 2 rt. angles.
\therefore PR || QS & PQ || RS (given).
\therefore PQSR is a parallelogram.
\therefore PQ = RS Proved.
Question 8
I drew a triangle ABC of which AB = AC and E is any point on BC produced. If the circumcircle of ABC intersects AE at the point D. Let us prove that ∠ACD = ∠AEC.
Solution:
In ∆ABC, AB = AC, E is a point on produced BC.
AE, cuts the circumcircle of ∆ABC at D.
To prove, ∠ACD = ∠AEC. Join C, D.
Proof : ABCD is a cyclic quadrilateral. \therefore ∠ABC + ∠ADC = 2 r. angles.
Again, straight line CD meets AE at D.
\therefore ∠ADC + ∠CDE = 2 rt. angles.
\therefore ∠ABC + ∠ADC = ∠ADC + ∠CDE
\therefore ∠ABC = ∠CDE
In ∆ABC, AB = AC, \quad\therefore ∠ABC = ∠ACB
\therefore ∠ACB = ∠CDE
Again, side EC of ∠CDE is produced.
\therefore External ∠DCB = ∠CDE + ∠CED
or, ∠ACB + ∠ACD = ∠CDE + ∠AEC
or, ∠CDE + ∠ACD = ∠CDE + ∠AEC[\because ∠ACD = ∠CDE]
\therefore ∠ACD = ∠AEC. Proved.
Question 9
ABCD is cyclic quadrilateral. Chord DE is the external bisector of ∠BDC. Let us prove that AE (or produced AE) is the external bisector of ∠BAC.
Solution:
ABCD is a cyclic quadrilateral, chord DE is the external bisector of ∠BDC.
To prove AE (or produced AE) is the external bisector of ∠BAC.
Proof : DE is the external bisector of ∠BDC.
\therefore ∠AED = ∠ACD = ∠ABC (angles on the same segment)
Again, ∠BAC = ∠BDC (angles on the same segment) [as, DE is the external bisector of ∠BDC]
\therefore A E is the external bisector of ∠BAC.
Question 10
BE and CF are perpendicular on sides AC And AB of triangle ABC respectively. Let us prove that four points B, C, E, F are on the same circle.
Solution:
In ∆ABC, BE & CF are two perpendiculars on side AC & AB.
To prove points B, C, E, F are on the same circle.
From that prove also ∆AEF & ∆ABC Are equiangular.
Join E, F.
Proof : AS ⊥ BE ⊥ AC & CF ⊥ AB
\therefore ∠BEC = 1 rt angle & ∠BFC = 1 rt. angle
\therefore ∠BEC = ∠BFC = 1 rt. angle
\therefore Points B, C, E, F are on the same circle.
Now, in ∆AEF & ∆ABC,
∠FAE = ∠BAC (Same angle)
Question 11
ABCD is parallelogram. A circle passing through the points A and B intersects the sides AD and BC At the points E and F respectively. Let us prove that the fou points E, F, C, D are concyclic.
Solution:
ABCD is a parallelogram. A circle passing through the points A & B intersects the sides AB & BC At E & F respectively.
To prove that the points E, F, C, D are concyclic Join E, F.
Proof : ABCD is a parallelogram.
\therefore ∠BAD + ∠ADC = 2 rt. angles.
Again, ABEF is a cyclic quadrilateral.
\therefore ∠BAE + ∠BFE = 2 rt. angles
But ∠BFE + ∠C FE = 2 rt. angles.
\therefore ∠EFC = ∠BAE \quad \therefore ∠EFC = ∠BAD
\therefore ∠EFC + ∠ADC = 2 rt. angles.
∠EFC + ∠EDC = 2 r. angles.
\therefore ∠EFCD is a cyclic quadrilateral.
\therefore Points E, F, C, D are concyclic
Question 12
ABCD is a cyclic quadrilateral. The two sides AB and DCAre produced to meet at the point P and the two sides AD and BC Are produced to meet at the point R. The two circumcircles of ∆BCP and ∆CDR intersect at the point T. Let us prove that the points P, T, R are collinear.
Solution:
To prove the points P, T, R are collinear.
Joint P, T ; T, R & C, T.
In BCP & ADP,
∠BPC = ∠APD ; ∠PBC = ∠ADP
& ∠BCP = ∠PAD
\therefore ∆BCP & ∆APD are equiangular.
i.e., they are similar.
Now, in ∆PCT & ∆RCT,
∠PCT = ∠CRT,
CT common & ∠RCT = ∠TPC
\therefore ∆PCT ≅ ∆RCT
\therefore ∠PTC = ∠RTC (corresponding angle)
As CT is the common side of ∠PTC & ∠RTC
\therefore TR & TP are on same straight line.
\therefore P, T, R are collinear.
Question 13
If point O is the ortho of triangle ABG, let us prove that it is the incentre of ∆DEF.
Solution:
To prove O is the incentre of the ∆DEF. In ∆BDO & ∆AEO.
∠BOD = ∠AOE, ∠BDO = ∠AEO = 90°
& Rest ∠BDO = Rest ∠EAO
\therefore ∆BDO ≅ ∆AEO,
\therefore OD = OE
Similarly, ∆BOD ≅ ∆AFO
\therefore OD = OF
\therefore OD = OE = OF
O is the incentre of the ∆DEF.
Question 14
I drew a cyclic quadrilateral ABCD such that AC bisected ∠BAD. Now produced AD to a point E is such away that DE = AB. Let us prove that CE = CA.
Solution:
To prove CE = CA
Join B, D.
Proof : Diagonal AC bisects ∠BAD.
\therefore ∠BAC = ∠DAC
Again, ∠DBC = ∠DAC
As these angles are on the same segment DC,
Again, ∠BDC = ∠BAC
As on the same segment BC,
\therefore ∠DBC = ∠BDC
\therefore BC = DC .
Again, ∠ACD = ∠ABD
As on the same segment AD,
\therefore ∠ABC = ∠ABD + ∠DBC
∠CDE = ∠DAC + ∠ACD
\therefore ∠ABC = ∠CDE
Now, in ∆ABC & ∆CED,
AB = DE (given), BC = DC & ∠ABC = ∠CDE
\therefore ABC ≅ ∆CED (SAS)
\therefore AC = CE
Question 15
In two circles, one circle passes through the centre Q of the other circle and the two circles meet each other at the point A intersect the circle which passes through the point O at the point P and the circle with centre O at the point R. Joining P, B and R, B. Let us prove that PR = PQ.
Solution:
Let the straight line PB cuts the circle with centre O at the point C.
Join A, C.
In ∆PRB & ∆ACP
∠BPR = ∠APC (same angle)
∠PBR = ∠CAP[ \because ARBC is a cyclic quadrilateral]
\therefore ∠PBR + ∠RAC = 2 rt. angles
= ∠RAC + ∠CAP
\therefore ∆PRB + ∆ACP are obtuse angled triangle, i.e., similar
\therefore ∠PRB = ∠PBR
\therefore PB = PR Proved.
Question 16
Let us prove that any four vertices of a regular pentagon are concyclic.
Solution:
Let ABCDE is a pentagon.
To prove A, B, C, E are concyclic.
Join C, E parallel to A B.
As AB || CE
\therefore ∠ABC + ∠EAB = 2 rt. angles
Again, ∠ABC + ∠BCE = 2 rt. angles
As sum of opposite angles of the quadrilateral
ABCE are supplementary.
\therefore Points A, B, C, E are concyclic.
17. Very Short Answer type questions (V.S.A.) :
M.C.Q. :
Question 1
In the picture beside O is the centre of circle and AB is a diameter. ABCD is a cyclic quadrilateral. ∠BAC is
- 50°
- 60°
- 30°
- 40°
Solution:
ABC = 180° – 120° = 60°
\therefore ∠ACB = 90° (semicircle angle)
∠BAC = 180° – (90° + 60°) = 180° – 150° = 30°
Ans. 30°
Question 2
In picture beside O is the centre of the circle and AB is a diameter. ABCD is a cyclic quadrilateral. The value of ∠BCD is
- 75°
- 105°
- 115°
- 80°
Solution:
∠ACB = 90° (semicircle angle)
∠ADC = 180° – 65° = 115°
∠ACD = 180° – (115° + 40°) = 180° – 155° = 25°
∠BCD = (90° + 25°) = 115°
Ans. 115°
Question 3
In picture beside O is the centre of circle and AB is diameter. ABCD is a cyclic quadrilateral in which AB || DC and if ∠BAC = 25° then the value of ∠DAC is
- 50°
- 25°
- 130°
- 40°
Solution:
∠ACB = 90° (semicircle angle)
\therefore ∠ABC = 180° – (90° + 25°) = 180° – 115° = 65°
∠ADC = 180° – 65° = 115°
∠DCA = alternate ∠C A B = 25°
∠DAC = 180° – (115° + 25°) = 180° – 140° = 40°
Ans. 40°
Question 4
In picture beside ABCD is a cyclic quadrilateral. BA is produced to the point F. If AE || CD, ∠ABC = 92° and ∠FAE = 20°, then the value of ∠BCD is
- 20°
- 88°
- 108°
- 72°
Solution:
∠ADC = 180° – 92° = 88°
∠DAF = ADC = 88°[\because AE || CD]
∠DAE = 88° + 20° = 108°
∠BAD = 180° – 108° = 72°
∠BCD = 180° – 72° = 108°
Ans. 108°
Question 5
In picture beside two circles intersect each other at the points C and D. Two straight lines through the points D and C intersect one circle at the points E and F respectively. If ∠DAB = 75°, then the value of ∠DEF is
- 75°
- 70°
- 60°
- 105°
Solution:
Join D, C.
∠DCF = ∠BAD = 75°
∠DEF = 180° – 75° = 105°
Ans. 105°
(B) Let us write true or false :
Question 1
The opposite angles of a cyclic quadrilateral are complementary.
Solution:
FALSE
Question 2
If any side of a cyclic quadrilateral be produced, the exterior angle so formed is equal to the interior opposite angle.
Solution:
TRUE
(C) Let us fill in the blanks :
Question 1
If the opposite angles of a quadrilateral be supplementary then the vertices of the quadrilateral will be __________________.
Solution:
Concyclic.
Question 2
A cyclic parallelogram is a ________________________ picture.
Solution:
Rectangle.
Question 3
The vertices of a cyclic square are ______________________.
Solution:
Concyclic.
Short Answer type questions :
Question 1
In the picture beside two circles with centres P are Q intersect each other at the points B and C. ACD is a line segment. If ∠ARB = 150°, ∠BQD = x°, then let us find the value of x.
Solution:
∠BCD = 150°
\therefore Reflex ∠BQD = 2 × 150° = 300°
\therefore x° = ∠BQD = 360° – 300° = 60°
\therefore x = 60°
Question 2
In picture beside two circles intersect at the points P and Q. If ∠QAD = 80° and ∠PDA = 84°, then let us find the value of ∠QBC and ∠BCP.
Solution:
∠QPC = ∠DAQ = 80°
∠QBC = 180° – 80° = 100°
∠PQB = ∠ADP = 84°
∠BCP = ∠AQD = 180° – (100° + 60°) = 20°
Question 3
In picture beside if ∠BAD = 60°, ∠ABC = 80°, then let us find the value of ∠DPC and ∠BQC.
Solution:
∠ADQ = ∠ADC = 180° – 80° = 100°
∠DPC = ∠APB = 180° – (80° + 60°) = 40°
∠BQC = ∠AQD = 180° – (100° + 60°) = 20°
Question 4
In picture beside O is the centre of the circle and AC is its diameter. If ∠AOD = 140° and ∠CAB = 50°, let us find the value of ∠BED.
Solution:
∠BOC = 180° – 80° = 100°
∠BEC = \frac{1}{2} ∠BOC = \frac{1}{2} × 100° = 50°
\therefore OB = OC
∠OCB = ∠O B C = \frac{180° - 100°}{2} = \frac{80°}{2} = 40°
∠BCF = (40° + 10°) = 50° = ∠CED
\therefore ∠BED = (50° + 50°) = 100°
Question 5
In picture beside O is the centre of the circle and AB is its diameter. If ∠AOD = 140° and ∠CAD = 50°, let us find the value of ∠BED.
Solution:
∠BOD = 180° – 140° = 40°
\therefore OB = OD
∠OBD = ∠ODB = \frac{140°}{2} = 70°
∠DBE = 180° – 70° = 110°
∠ACF = 180° – 70° = 110°
∠BED = ∠AEC = 180° – (110° + 50°) = 180° – 60° = 20°