OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Ex 14(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(b)

Question 1.
In the fig., in △ABC, AB = AC, and XY || BC. Prove that BCYX is a cyclic quadrilateral.

Solution:
Given : In the figure, in △ABC, AB = AC. XY || BC
To prove : BCYX is a cyclic quadrilateral
Proof : In △ABC,
XY || BC
∴∠AXY = ∠ABC (Corresponding angles)
But △ABC = ∠ACB (∵AB = AC)
∴∠AXY = ∠ACB
But Ext. ∠AXY is equal to its interior opp. ∠ACB
∴ BCYX is a cyclic quadrilateral Hence proved.

Question 2.
In the figure, In a quad. ABCD, ext. ∠XCD = int. opp ∠A. Prove that the quad. ABCD is a cyclic quad.

Solution:
Given : In the figure, in quad. ABCD
BC is produced to X
Ext. ∠XCD = Int. opp. ∠A
To prove : Quad. ABCD is a cyclic
Proof: ∠XCD + ∠DCB = 180° (Linear pair)
∠A = ∠DCB = 180° (∵ ∠A = ∠XCD)
But these are opposite angles of a quad.
∴ Quad ABCD is a cyclic Hence proved.

Question 3.
In the figure, PQR is an isosceles triangle with PQ equal to PR. A circle passes through Q and R and intersects the sides PQ and PR at points S and T respectively. Prove that QR || ST.

Solution:
Given : In △PQR, PQ = PR
A circle passing through Q and R, intersects PQ and PR at S and T respectively. ST is joined
To prove : ST || QR
Proof: In △PQR
PQ = PR (given)
∴ ∠Q = ∠R ….(i) (Angles opposite to equal sides)
∵ SQRT is a cyclic quadrilateral
∴ Ext. ∠S = Int. opp. ∠R …(ii)
From (i) and (ii)
∠Q = ∠S
But these are corresponding angles
∴ QR || ST Hence proved.

Question 4.
In the figure, AB is the common chord of two circles. If AC and AD are diameters, prove that D, B and C are in a straight line. O₁ and O₂ are the centres of the circles.
Solution:

Given : Two circles with centres O₁ and O₂
intersect each other at A and B
AO₁C and AO₂D are diameters
To prove : D, B and C are in a straight line or D, B and C are collinear
Proof : ∵ AC is the diameter of circle of centre O₁
∴ ∠ABC = 90° (Angle in a semicircle)
Similarly
∠ABD = 90° (Angle in a semicircle)
Adding we get,
∠ABC + ∠ABD = 90° + 90° = 180°
∴ CBD is a straight line
Hence D, B and C are in the same straight line Hence proved.

Question 5.
In figure, AB is the diameter of the circle whose centre is O. AD and BC are perpendiculars to the line XY. CB meets the circle at E. Prove that CE = AD.

Solution:
Given : In a circle with centre O, AB is its diameter AD ⊥ XY and BC ⊥ XY which intersects the circle at E
To prove : CE = AD
Construction : Join A, E.

Proof: ∠AEB = 90° (Angle in a semicircle)
∴ ∠AEC = 90°
(∵ ∠AEB + ∠AEC = 180° linear pair)
But ∠C = ∠D = 90°
(∵ AD and EC are perpendicular to XY)
∴ AECD is a rectangle
∵ Opposite sides of a rectangle are equal
∴ CE =AD Hence proved.

Question 6.
In fig., AB and CD are parallel chords of a circle whose diameter is AC. Prove that AB = CD.

Solution:
Given : In a circle with centre O, AC is its diameter and chord AB || CD
To prove : AB = CD
Construction : Join OB and OD

Proof: In AAOB and ACOD,
OA = OC
OB = OD (radii of the same circle)
∠BAO = ∠OCD (Alternate angles)
∴ △AOB ≅ △COD (SSA axiom)
∴ AB = CD Hence proved.

Question 7.
In figure, APB and CQD are straight lines through the points of intersection of two circles. Prove
(i) AC || BD,
(ii) ∠CPD = ∠AQB (SC)

Solution:

Given : Two circles intersect each other at P and Q
Lines APB and CQD are drawn from the point of intersection respectively PQ, AC, BD and AQ, QB, CP and PD are joined
To prove :
(i) AC || BD (ii) ∠CPD = ∠AQB
Proof:
(i) ∵ APQC is a cyclic quad.
∴ Ext. ∠BPQ = Int. opp. ∠C ….(i)
∵ PBDQ is a cyclic quad.
∴ ∠BPQ + ∠D = 180° (sum of opp. angles)
⇒ ∠C + ∠D = 180° {from (∠)}
But these are co-interior angles
∴ AC || BQ

(ii) In △AQB and △CPD,
∠PAQ = PCQ (Angles in the same segment)
∠PBQ = ∠PDQ (Angles in the same segment)
or ∠BAQ = ∠ABQ
∠ABQ = ∠PDC
∴ △AQB ~ △APD (AA axiom)
∴ Third angle = Third angle
⇒ ∠AQB = ∠CPD Hence proved.

Question 8.
Prove that the circle drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.
Solution:
(a) Given: ABCD is a rhombus whose diagonals AC and BD intersect each other at O. A circle with AB as diameter is drawn.

To prove : The circle passes through O
Proof : Let the circle drawn on AB as diameter does not passes through O, let it intersect AC at P
Join PB
The ∠APB = 90° (Angle in a semicircle)
But ∠AOB = 90°
(∵ The diagonals bisect each other at right angles)
∴ ∠APB = ∠AOB
But it is not possible because ∠APB is the exterior angle of AOPB and an exterior angle of a triangle is always greater than its interior opposite angle
∴ Our supposition is wrong
Hence the circle will pass through O Hence proved.

Question 9.
If two non-parallel sides of a trapezium are equal, it is cyclic.
OR
An isosceles trapezium is always cyclic.

Solution:
Given : An isosceles trapezium ABCD in
which AD || BC and AB = DC
To prove : ABCD is cyclic
Construction : Draw AE and DF perpendicular on BC
Proof: In right △ABE and △DCF
Hyp. AB = DC (given)
Side AE = DF (Distance between two parallel lines)
∴ △ABE ≅ △DCF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now AD || BC
∴ ∠DAB + ∠B = 180° (Co-interior angles)
⇒ ∠DAB + ∠C = 180° (∵ ∠B = ∠C proved)
But these are sum of opposite angles of a quad.
∴ ABCD is a cyclic. Hence proved.

Question 10.
Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

Solution:
Given : PQRS is a cyclic quadrilateral
∠A, ∠B, ∠C and ∠D are angles in the four segment so formed exterior to the cyclic quadrilateral
To prove : ∠A + ∠B + ∠C + ∠D = 6 right angles
Construction : Join AS and AR
Proof: In cyclic quad. ASDP,
∠PAS + ∠D = 2 rt. angles ….(i) (Sum of opposite angles)
Similarly in cyclic quad. ARBQ,
∠RAQ + ∠B = 2 rt. angles ….(ii)
and in cyclic quad. ARCS,
∠SAR + ∠C = 2rt. angles ….(iii)
Adding (i), (ii) and (iii)
∠PAS + ∠D + ∠RAQ + ∠B + ∠SAR + ∠C = 2 + 2 + 2 = 6rt. angles
⇒ PAS + ∠SAR + ∠RAQ + ∠B + ∠C + ∠D = 6 rt. angles
⇒ ∠A + ∠B + ∠C + ∠D = 6 rt. angles Hence proved.

Question 11.
(i) In figure, O is the circumcentre of △ABC and OD ⊥ BC. Prove that ∠BOD = ∠A.

Solution:
Given : O is the circumcentre of △ABC
OD ⊥ BC. OB and OC are joined
To prove : ∠BOD = ∠A
Proof: Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC ….(i)
In △OBC, OD ⊥ BC
OB = OC (radii of the same circle)
∴ OD bisects ∠BOC
⇒ ∠BOD = \(\frac { 1 }{ 2 }\) ∠BOC ….(ii)
From (i) and (ii)
∠BOD = \(\frac { 1 }{ 2 }\) ∠BAC = ∠BAC
Hence ∠BOD = ∠A
Hence proved.

Question 12.
ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

Solution:
Given : ABCD is a cyclic quadrilateral
A circle passing through A and B meet AD
and BC at E and F respectively
EF is joined
To prove : EF || DC
Proof: ∵ ABCD is a cyclic quad.
∴ ∠1 + ∠3 = 180° ….(i) (Sum of opposite angles)
Similarly ABFE is a cyclic quadrilateral
∴ ∠1 + ∠2 = 180° ….(ii)
From (i) and (ii)
∠1 + ∠3 = ∠1 + ∠2 ⇒ ∠3 = ∠2
But these are corresponding angles
∴ EF || DC Hence proved.