Mathematics [4 marks sum]

To proof:- QT = TR
Proof: Let the circle touches sides PQ and PR at points A and B respectively.
PA = PB , AQ = QT and BR = TR .....(Lengths of tangents drawn from an external point to a circle are equal)
Given, PQ = PR
PA + AQ = PB + BR
AQ = BR {Using (1))
⇒ QT = TR

Let the sides of parallelogram ABCD touch the circle at points P, Q, R and S.
AP = AS - (1)
PB = BQ - {2} {Length of tangents drawn from an external point to a circle a equal)
DR = DS - {3}
RC = CQ - (4)
Adding (1), {2}, {3} and (4)
AP + PB + DR + RC = AS + BQ + DS + CQ
AB + CD = AD + BC
2 AB = 2 BC => AB = BC {Opposite sides of a parallelogram are equal)
:. AB = BC = CD = DA,
Hence , ABCD is a rhombus.

In Δ AOP =Δ BOP
AP = PB (lengths of tangents drawn from and external point to a circle are equal)
OP = PO (common)
∠ PAO = ∠ PBO = 90 ° (radius is ⊥  to tangent at the point of contact)
∴  Δ AOP ≅  Δ BOP   {By RHS}
Δ AOP =  Δ BOP {By CPCT}
In Δ AMO and Δ BMO
AO = OB (radius of same circle)
∠ MOA = ∠ MOB {Proved above)
OM = MO {Common)
∴ Δ AMO ≅  Δ BMO {By CPCT}
∠ AMO = ∠ BMO
∠ AMO + ∠ BMO = 180°
∴ 2 ∠ AMO = 180°
∠ BMO = ∠ AMO =  90°
Hence, OP is the perpendicular bisector of AB. 

PA = PB = 20 Units ---(1) {Length of tangents drawn from an external point to a circle are equal }
CQ = CA and DQ = DB
Perimeter of Δ PCD
= PC+ CD+ PD
= PC+ CQ + QD + PD {Using {1)}
= PA+ PB
= 2PA
= 2{20}
= 40 Units

∠ NYB =50 °
∠ YNB = 20°
In  Δ NYB,
20° + ∠ NBY + 50 ° = 180°
⇒ ∠ NBY = 180°  - 70 ° = 110 °
Now, ∠ MAN = ∠ NBM = 110°  (Angles in the same segment)
∠ MON = 2 ∠ MAN (Arc MN subtends ∠ MON at centre and ∠ MAN at remaining part of the circle)
∠ MON = 2(110°) = 220°
Reflex ∠ MON = 360° - ∠ MON
= 360° - 220°
Reflex ∠ MON = 140°. 

If two chords of a circle interest internally then the products of the lengths of segments are equal, then
AP x BP= CP x DP ... ( 1)
But, AP= CP (Given) ....(2)
Then from ( 1) and (2), we have
BP= DP ......(3)
Adding (2) and (3),
AP + BP= CP + DP
⇒ AB = CD
Hence Proved.

Arc AC subtends LAOC at the centre of circle and LABC on the circumference of the cirde .
∴ ∠ AOC = 2 ∠ ABC ...(1)
Similarly, ∠ BOD and ∠ DCB are the angles subtended by the arc DB at the centre and on the circumference of the circle respectively .
∴ ∠ BOD = 2 ∠ DCB ... (2)
Adding ( 1) and (2),
∠ AOC+ ∠ BOD = 2(∠ ABC + ∠ DCB) ... (3)
In triangle ECB ,
∠ AEC = ∠ ECB + ∠ EBC = ∠ DCB + ∠ ABC
From (3),
∠ AOC+ ∠ BOD = 2 ∠ AEC
Hence Proved.

m ∠ A = 3 (m ∠C)
∠ A + ∠ C = 180 (Opposite angles of a cyclic quadrilateral)
3∠C + ∠ C = 180
4 ∠ C = 180
∠ C = 45
m ∠ A = 3 (m ∠ C)
= 3 × 45
= 135
m ∠ A = 135°

Let O be the centre of the cirde on diameter AC of the circle
Since, EC make ∠ EOC at the centre and ∠ EBC on the remaining part of the circle
∴ ∠ EOC = 2 ∠EBC = 2 (65) = 130°
In Δ EOC ,
∠EOC + ∠ OCE + ∠ CEO =  180°
130 + x + x = 180°  (OE = OC , ∴ ∠ OEC = ∠ OCE = x)
2x = 50
x = 25
∠ OCE = ∠ OEC  = 25°
Also , ∠ OCE = ∠ CED = 25°   (alternate interior angles)

In cyclic quadrilateral ABCD
∠A + ∠ C = 180°
1/2 ∠A + 1/2 ∠C = 90°
∠EAB + ∠BCF = 90° -(1) (AE bisects ∠ A ; CF bisects ∠C)
Also ,
∠BCF = ∠BAF - (2) (Angles in the same segment)
Using (1) in (2) we get ,
∠EAB + ∠BAF = 90°
∠FAE = 90°
EF is the diameter of the circle ,
∴ angle in a semi circle is a right angle.

To prove = DE II BC
Proof: In cydic quadrilateral DECB
∠ DEC + ∠ DBC = 80° - ( 1) (Opposite angles of cyclic quadrilateral)
Also, ∠ AED + ∠ DEC = 80° - (2) (Linear pair)
From (1) and (2), we get,
∠ DBC = ∠ AED - (3)
AB= AC (given)
∴ ∠ABC = ∠ ACB - (4) (angles opposite to equal sides of triangle)
From (3) and ( 4) ⇒ ∠ AED = ∠ ACB
But, these are corresponding angles
∴ DE II BC

We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠ APO =90° - (1)
Also, AD is the diameter of the circle with centre 0
∴ ∠ APD=90° - (2) (Angle in semi circle)
From ( 1) and (2), we get, The cirde drawn with any side of a rhombus as a diameter, passes through point of intersection of its diagonals.

To prove: BD =DC
Proof: let .AB be the diameter of the circle with centre O
∴ ∠ ADB = 900 (Angles in a semicirde is a right triangle)
∠ ADB + ∠ ADC = 180 (linear pair)
∴ ∠ ADC = 180 - 90 = 90°
In Δ ADB and Δ ADC
AB= AC (given)
∠ ADB = ∠ ADC (90"each)
AD= AD (Common)
∴ Δ ADB ≅ Δ ADC (RHS)
Hence BD = DC (CPCT)

Given: Two congruent circles with centre 0 and P. Mis the mid-point of OP
To prove: Chord AB and CD are equal.
Construction: Draw OQ ⊥ AB and PR ⊥ CD.
Proof: In Δ OQM and Δ PRM
∠ OQM = ∠ PRC ...(Each 90°)
OM =MP ....(As M is the mid-point)
∠OMQ = ∠ PMR ...(Verically opposite angles)
Therefore, Δ OQM ≅ ΔPRM
⇒ OQ = PR ...(By CPCT)
Now the perpendicular distances of two chords 1n two congruent circles are equal, therefore chords are also equal.
⇒ AB = CD
Hence proved.

AM = MB
CN = ND
∴ OM ⊥ AB
and ON ⊥ CD
(A line bisecting the chord and passing through the centre of the circle is perpendicular to the chord)
∴ ∠ OMA = ∠ OND = 90° each
But these are alternate interior angles
∴ AB || CD

Let QT be the diameter of ∠ PQR
Since , PQ = QR
∴ OM = ON
In Δ OMQ and Δ ONQ
OM = ON (equal chords are equidistant from the centre)
∠ OMQ = ∠ ONQ (90 ° each)
OQ = OQ (common)
Δ OMQ ≅ Δ ONQ (RHS)
∴ ∠ OQM = ∠ OQN (CPCT)
Thus QT i.e. diameter of the circle visects ∠ PQR.

Draw perpendiculars OR and OS to CD and AB respectively.
In triangle ORP and triangle OSP
OP= OP
OR = OS (Distance of equal chords from centre are equal)
∠ PRO = ∠ PSO (right angles)
Therefore, Δ ORP ≅ Δ OSP
Hence, ∠ RPO = ∠ SPO
Thus OP bisects ∠ CPB.

Join OX and OZ
In Δ XOY and Δ ZOY
OX = YZ (radii of same circle)
XY = YZ (given)
OY = OY (common)
∴ Δ XOY ≅ Δ ZOY (SSS)
∴ ∠ OYX = ∠ OYZ (CPCT)
Hence , OY is the bisector of ∠XYZ passing through O.

Here, ABCD is a cyclic quadrilateral. PM is bisector of ∠ APB and QM is bisector of ∠ AQD
In Δ PDL and Δ PBN, ∠ 1=∠ 2 (PM is a bisector of LP)
∠ 3 = ∠ 9 (exterior angle of cydic quad. = interior opposite angle)
∴ ∠ 4 = ∠ 7
But, ∠ 4 = ∠ 8 (vertical opposite angles)
∴ ∠ 7 = ∠ 8
Now in Δ QMN and Δ QML
∠ 7 = ∠ 8 (proved above)
∠ S = ∠ 6 (QM is bisector of Q)
∴ Δ QMN ~ Δ QML
⇒ ∠ QMN and ∠ QML
But, ∠ QMN + ∠ QML = 180°
∴ ∠ QMN = ∠ QML = 90°
Hence, ∠ PMQ = 90 °

Given: Two cirdes with centres 0 and P, and MN II OP || RQ
To prove: (i) MN = 20P (ii) MN= RQ.
Construction: OX ⊥ MN, PY ⊥ MN, OW ⊥ RZ, PZ ⊥ RQ
Proof: Since each angle of the quadrilateral XYZW is a right angle, sc XYZW is a rectangle.
Also, XYPO is a rectangle. ...(1)
Now, perpendicular drawn from the centre to the chord bisects the chord.
Therefore, MA = 2 XA and AN = 2 AY ...(2)
Now, MN = MA + AN = 2 XA + 2 AY [from (2)]
⇒ MN = 2(XA + AY) = 2 XY
⇒ MN = 2 OP [As XYPO is a rectangle, XY = OP] ... (3)
This proves part (i).
By similar arguments, we have RQ = 2 OP ...(4)
Using (3) and ( 4), we get
MN= RQ.
This proves part (ii).

Given: AB and CO are two chords of a cirde with centre O.
AB II CD , M and N are midpoints of AB and CO respectively.
To prove : MN passes through centre O.
Construction : Join OM, ON, and through O, draw a straight line EF parallel to AB.
Proof : OM ⊥ AB
(line joining the midpoin t of a chord to the centre of a circle is perpendicular to it)
∠ AMO = 90°
∠ MOE = 90° [cointerior angle of ∠ AMO]
∠ NOE = 90° [corresponding angle of ∠ AMO]
∠ MOE + ∠ NOE = 180°
∠ MON is a straight line .
Hence, MN passes through centre O.

Given: AB = AC, ∠ ABO = ∠ CBO
To Prove: AB = BC
Construction : Draw ON ⊥ AB and OM ⊥ BC
Proof : In triangles BNO and BMO,
∠ NBO = ∠ MBO (Given)
∠ BNO = ∠ BMO (Each 90 ° )
BO= BO (common)
Thus , Δ BNO ≅ Δ BMO (By AAS)
⇒ BN =BM
⇒ 2 BN = 2 BM (Since perpendicular drawn from the centre bisects the chord)
⇒ AB = BC
Hence Proved.

M and N are mid points of equal diords AB and CD respectively.
ON ⊥ CD and OM ⊥ AB
∴ ∠ ONC = ∠ OMA (90° each) ...(1)
(A line bisecting the chord and passing through the centre of the circle is perpendicular to the chord)
∴ AB = CD
ON = OM (equal chords are equidistant from the centre)
In Δ MON ,
MO = NO
∴ ∠ ONM = ∠ OMN ..(2)
Subtracting (2) from ( 1)
∠ONC - ∠ ONM = ∠ OMA - ∠ OMN
∠ CNM =∠ AMN