Chapter – 10 : Theorems Related To Cyclic Quadrilaterial

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