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Wednesday, January 26, 2022
हार्दिक स्वागतम
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The Greek Mathematician Euclid, who lived around 300 BC, wrote Elements, a 13-volume work on the Principals of geometry and properties of numbers. The plane geometry is discovered by him therefore he is considered as father of Geometry. In this book he had written “there is no royal road to geometry”. His work was rediscovered in the 15th century, when it was translated from Arabic, and until recent years has been the principal source for the study of geometry.
EUCLID
Father of Geometry
Geometry-Circle
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Definition of terms related to circle
Arc
Circumference
Radius
Diameter
Semicircle
Chord
Concentric circles
Segment
Intersecting circles
Secant of a circle
Sector
�
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Arc(चाप)
D
A
B
C
O
NRJ
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Circumference(परिधि)
Definition: The total length of boundary line of a circle is called circumference.
(कुनै पनि बृत्तको बाहिरी घेराको जम्मा लम्बाईलाई त्यस बृत्तको परिधि भनिन्छ।)
Circumference
NRJ
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Radius(अर्धब्यास)
Definition: It the line segment that joins the center of a circle and any point on its circumference. It is denoted by (r).
(बृत्तको केन्द्रबिन्दु र परिधिको कुनै एक बिन्दु जोड्ने रेखाखण्डलाई त्यस बृत्तको अर्धब्यास भनिन्छ। यसलाई r ले जनाईन्छ।)
In the given figure, OA is the radius.
(चित्रमा OA अर्धब्यास हो।)
OA = r
A
O
NRJ
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Diameter
(ब्यास)
Definition: A line segment that passes through the center of a circle and joins any two points on its circumference is said to be Diameter of the circle. The length of a diameter of a circle is always twice of its radius.
(बृत्तको केन्द्रबिन्दु भएर जाने र परिधिका कुनै दुई बिन्दुहरुलाई जोड्ने रेखाखण्डलाई त्यस बृत्तको ब्यास भनिन्छ। ब्यासको लम्बाई जहिले पनि अर्धब्यासको दुई गुणा हुन्छ।)
In the Figure , AOB is the diameter.
(चित्रमा, रेखाखण्ड AOB ब्यास हो।)
O
B
A
NRJ
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Semi-circle
अर्ध-बृत्त
Definition: A diameter divides a circle into two equal halves and each half is called semi- circle.
(ब्यासले कुनैपनि बृत्तलाई दुई बराबर भागमा बिभाजन गर्दछ र ती प्रत्येक भागहरुलाई अर्ध-बृत्तहरु भनिन्छ।)
In the figure, ACB and ADB are Semi circle.
(चित्रमा, ACB र ADB अर्ध-बृत्तहरु हुन।)
NRJ
B
A
O
C
D
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Chord(जीवा)
Definition: The line segment that joins any two points on the circumference of a circle is the called the chord of the circle.
(बृत्तको परिधिको कुनै दुई बिन्दु जोड्ने रेखाखण्डलाई जीवा भनिन्छ।)
Note: Diameter is the longest chord of the circle.
(नोट: ब्यास नै बृत्तको सबैभन्दा लामो जीवा हो।)
In the figure, AB and PQ are chords.
Also CD is longest chord or diameter.
(चित्रमा, AB र PQ जीवाहरु हुन साथै CD सबै भन्दा लामो जीवा अर्थात ब्यास हो।)
NRJ
Q
P
O
B
A
C
D
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Concentric circles
(एक केन्द्रित बृत्तहरु)
Definition: Two or more circles are said to be concentric circles if they have the same Centre but different radii . In the figure given below are the Concentric circles.
(एउटै केन्द्रबिन्दु तथा फरक फरक अर्धब्यास भएका दुई वा दुई भन्दा बढि बृत्तहरुलाई एक केन्द्रित बृत्तहरु भनिन्छ। तलका दिईएका चित्र एक केन्द्रित बृत्तहरु हुन।)
NRJ
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Segment
NRJ
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Intersecting Circles
Definition: If two circles are intersect to each other at two points , they are said to be4 intersecting circles.
In the given figure, two circles are intersect at points Q and R.
R
Q
NRJ
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Tangent to a circle
Definition: a line that intersects the circle at exactly one point is called a tangent to the circle. The point at which the tangent intersects the circle is called the point of contact.
In the figure, NRJ is the tangent and R is the point of contact.
N
J
R
NRJ
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Secant of a circle
Definition: The Line is intersects a circle in two distinct/different points is called a secant of a circle.
In the figure, PQ is the secant of a circle.
P
B
O
A
Q
NRJ
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Sector
Definition: The region enclosed between any two radii and corresponding arc of a circle is called a sector of a circle.
In the given figure, AOBM is the minor sector
and AOBN is the major sector.
A
B
N
M
O
NRJ
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Important properties
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Theorem 1: The angle at the center of a circle is twice the angle at its circumference standing on the same arc.
C
A
O
D
B
Proof:
Alternative process
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S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between central angle and its opposite arc. |
2 | | 2 | Relation between inscribed angle and its opposite arc. |
3 | | 3 | From the statements (1) and (2). |
Proved.
C
A
B
O
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Theorem 2: The inscribed angle of a circle standing on the sane arc are equal.
Or, Angles on the same segment of a circle are equal.
S.N | Statements | S.N | Reasons |
1 | | 1 | Inscribed angle is half of central angle of the circle standing on the same arc. |
2 | | 2 | Same as reason (1) |
3 | | 3 | From statements (1) and (2) |
Proved.
D
A
O
B
C
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Alternative process
S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between inscribed angle and its opposite arc. |
2 | | 2 | Relation between inscribed angle and its opposite arc. |
3 | | 3 | From the statements (1) and (2). |
D
A
O
B
C
Proved.
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S.N | Statements | S.N | Reasons |
1 | | 1 | Inscribed angle is half of central angle of the circle standing on the same arc. |
2 | | 2 | Inscribed angle is half of central angle of the circle standing on the same arc. |
3 | | 3 | Adding statements (1) and (2). |
4 | | 4 | |
5 | | 5 | From statement (4). |
6 | | 5 | By joining O,B and O,D and using the above facts and reasons. |
Proof
A
D
C
B
O
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Unseen Theorem
R
P
Q
O
Proof
S.N | Statements | S.N | Reasons |
1 | | 1 | Straight angle |
2 | | 2 | Both are standing on the same arc. |
3 | | 3 | From statement (2) |
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Proof
O
P
Q
R
S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between central angle and inscribed angle. |
2 | | 2 | Same reason as (1). |
3 | | 3 | Adding statements (1) and (2). |
4 | | 4 | From statement (4) |
| | | Proved. |
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Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between central angle and inscribed angle. OR, Central angle is double of inscribed angle. Or, Both are standing on the same arc. |
2 | | 2 | |
3 | | 3 | Both are standing on the same arc. |
4 | | 4 | From statements (1), (2) and (3). |
5 | | 5 | From (4) and OA=OB. |
Proved.
O
C
A
B
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S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between Central angle and its opposite arc. |
2 | | 2 | Relation between Inscribed angle and its opposite arc. |
3 | | 3 | |
4 | | 4 | From statements (1) and (2). |
5 | | 5 | From (4) and corresponding angle are equal. |
Proof:
Proved.
D
A
B
O
C
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S.N | Statements | S.N | Reasons |
1 | | 1 | Given |
2 | | 2 | From statement (1). |
3 | | 3 | Relation between central angle and inscribed angle. |
4 | | 4 | From statements (2) and (3). |
5 | | 5 | From statement (4), corresponding angles are equal. |
D
A
B
O
C
Proof:
Proved.
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M
R
N
S
X
S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between inscribed angle and its opposite arc. |
2 | | 2 | Same reason as (1). |
3 | | 3 | Exterior angle of a triangle is equal to sum of its opposite two interior angles. |
4 | | 4 | From statements (1), (2) and (3). |
Proof:
Proved.
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S.N | Statements | S.N | Reasons |
1 | | 1 | Exterior angle of a triangle is equal to its opposite two interior angle. |
2 | | 2 | Inscribed angle and its opposite arc |
3 | | 3 | Same as reason (2). |
4 | | 4 | From statements (2) and (3). |
Proof:
D
F
X
E
G
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D
O
N
M
C
A
B
S.N | Statements | S.N | Reasons |
1 | | 1 | Given |
2 | | 2 | |
3 | | 3 | Equal arc subtend equal angles at circumference. |
4 | AM=BN | 4 | Corresponding chords of equal arc. |
5 | | 5 | From statements (1), (2), (3) and (4). |
6 | | 6 | Corresponding angle of a congruent triangle. |
Proof:
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S.N | Statements | S.N | Reasons |
1 | | 1 | Exterior angle of a triangle is equal to sum of its opposite two interior angles. |
2 | | 2 | Same reason as (1). |
3 | | 3 | Inscribed angle and its opposite arc. |
4 | | 4 | Same reason as (3) |
5 | | 5 | Same reason as (3) |
6 | | 6 | Same reason as (3) |
7 | | 7 | Both are right angles. |
8 | | 8 | From statements (1), (2) and (7). |
9 | | 9 | From statements(3,4,5,6 and 7) |
10 | | 10 | Solving statement 9. |
Proof
A
C
B
D
X
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Proof
A
C
B
D
X
S.N | Statements | S.N | Reasons |
1 | | 1 | Exterior angle of a triangle is equal to sum of its opposite two interior angles. |
2 | | 2 | Same reason as (1). |
3 | | 3 | Both are right angles. |
4 | | 4 | From statements (1) and (2). |
5 | | 5 | Multiply on both sides by 2 of statement (4). |
6 | | 6 | From statement (5) and double of inscribed angle is equal to its opposite arc. |
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S.N | Statements | S.N | Reasons |
1 | | 1 | Relation between central angle and its opposite arc. |
2 | | 2 | Same as reason (1). |
3 | | 3 | Base angle of an isosceles triangle. |
4 | | 4 | OD=OC or, Radii of the same circle. |
5 | | 5 | Ext. angle of a triangle is equal to sum of its opposite two interior angle. |
6 | | 6 | From statements (3) and (5). |
7 | | 7 | Ext. angle of a triangle is equal to sum of its opposite two interior angle. |
8 | | 8 | From statements (3) and (4). |
9 | | 9 | From statements (6) and (8). |
10 | | 10 | From statements (1,2 and 9). |
11 | | 11 | From statement 10 and 3 divided on both sides of st.10. |
Proof:
B
E
C
D
A
O
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S.N | Statements | S.N | Reasons |
1 | | 1 | |
2 | | 2 | Same arc BC adding on both sides of statement (1). |
3 | | 3 | From statement (2). |
4 | | 4 | From statement (4) and Both are made on equal arc. |
X
Y
A
B
C
D
Proof:
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S.N | Statements | S.N | Reasons |
1 | | 1 | Given. |
2 | | 2 | Same arc CD adding on both sides of statements 1. |
3 | | 3 | |
4 | | 4 | Both are standing on the equal arc Or, from statement 1.. |
5 | | 5 | Remaining angle of a triangle Or, Corresponding angle of a similar triangle. |
Proof:
A
C
E
D
B
X
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A
B
E
O
C
D
S.N | Statements | S.N | Reasons |
1 | | 1 | Both are standing on the same arc. |
2 | | 2 | Both are right angles. |
3 | | 3 | Common angles. |
4 | | 4 | Remaining angles of a triangles. |
5 | | 5 | From statements 1 and 4. |
Proof:
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S.N | Statements | S.N | Reasons |
1 | | 1 | Both are right angles. |
2 | | 2 | Common angles. |
3 | | 3 | Remaining angle of a triangles. |
4 | | 4 | OX=OY ( Radii of the same circle). |
5 | | 5 | From statements 3 and 4. |
Proof:
X
O
Z
Y
A
M
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Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Given. |
2 | | 2 | Both are standing on the same arc. |
3 | | 3 | Same reason as 2. |
4 | | 4 | From statements 1,2 and 3. |
5 | | 5 | From statement 4 and , alternate angle are equal. |
A
P
C
D
B
Q
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18. In the adjoining figure, POQ and ROT are two diameters of a circle with center at O. If Q is the midpoint of the arc TQS and ∡QOR is obtuse angle, prove that PQ∥RS .
Given: POQ and ROT are two diameters of a circle with center at O, Q is the midpoint of the arc TQS and ∡QOR is obtuse angle.
To prove: PQ∥RS.
Q
O
P
R
S
T
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Q is the mid-point of arc TQS. |
2 | | 2 | The central angle formed by equal arc. |
3 | | 3 | OR=OS (Radii of the same circle) |
4 | | 4 | Exterior angle of a triangle is equal to sum of its opposite two interior angle. |
5 | | 5 | From statements 2,3,and 4. |
6 | | 6 | From statement (5) and cancelled 2 from both side of statement (5) |
7 | PQ∥RS | 7 | From statement 6 and corresponding angle are equal. |
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A
D
C
E
B
S.N | Statements | S.N | Reasons |
1 | | 1 | Sum of linear pair. |
2 | | 2 | Opposite angle of a cyclic quadrilateral. |
3 | | 3 | From statements 1 and 2. |
4 | | 4 | From statements 3. |
Proof:
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20. ∆ABC is an isosceles triangle and XY∥BC. If XY cuts AB at X and AC at Y, Prove that the four points X,B,C and Y are concyclic.
A
X
Y
C
B
S.N | Statements | S.N | Reasons |
1 | | 1 | Base angle of an isosceles triangle. |
2 | | 2 | XY∥BC and corresponding angle. |
3 | | 3 | Same as reason 2. |
4 | | 4 | From statements 1,2 and 3. |
5 | X, B, C and Y are concyclic. | 5 | From statement 4 and exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
Given: ∆ABC is an isosceles triangle, XY∥BC And XY cuts AB at X and AC at Y.
To prove: X,B,C and Y are concyclic
Proof:
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21. In the given figure, QR is the bisector of ∡SQT and PQRS is a cyclic quadrilateral. Prove that ∆PSR is an isosceles triangle
P
S
R
Q
T
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | QR is the bisector of ∡SQT . |
2 | ∡SPR=∡SQTR | 2 | Both are standing on the same arc. |
3 | ∡PSR=∡RQT | 3 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
4 | ∡PSR=∡SPR | 4 | From statements 1,2 and 3. |
5 | PR=SR | 5 | From 4. |
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A
B
C
D
E
S.N | Statements | S.N | Reasons |
1 | | 1 | AC=BC |
2 | | 2 | Both are standing on the same base. |
3 | | 3 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle |
4 | | 4 | From statements 1, 2 and 3. |
5 | | 5 | From statements 3 and 4. |
6 | | 6 | From statement5. |
Proof:
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23. In the given figure, TPQ is an isosceles triangle in which TP=TQ. A circle passing through P and Q cuts TP and TQ at R and S respectively. Prove that PQ//RS.
P
Q
S
R
T
Given: Here, TPQ is an isosceles triangle in which TP=TQ, A circle passing through P and Q cuts TP and TQ at R and S respectively
To Prove: PQ//RS
Proof:
S.N | Statements | S.N | Reasons |
1 | TP=TQ | 1 | Given |
2 | | 2 | Base angle of an isosceles triangle. |
3 | | 3 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
4 | | 4 | From statements 2 and 3 |
5 | PQ//RS | 5 | From 4 and corresponding angle are equal. |
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24. ABCD is parallelogram. The circle through A, B and C intersect CD produced at E. Prove that AE=AD.
A
E
D
B
C
Given: ABCD is parallelogram and the circle through A, B and C intersect CD produced at E.
To prove: AE=AD
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Opposite angle of a parallelogram. |
2 | | 2 | Sum of opposite angle of a cyclic quadrilateral. |
3 | | 3 | Straight angle. |
4 | | 4 | From statements 2 and 3. |
5 | | 5 | |
6 | | 6 | |
7 | AE=AD | 7 | From statement 6. |
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25. n the given figure, PQRS is a cyclic quadrilateral. If the diagonals PR and QS are equal,
Prove that, QR=PS and PQ//SR.
S
P
Q
R
Given: PQRS is a cyclic quadrilateral and diagonals PR and QS are equal
To prove: QR=PS and PQ//SR
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Corresponding major arc of equal chords.(PR=QS) |
2 | | 2 | |
3 | PS=QR | 3 | Corresponding arc of equal arc. |
4 | | 4 | Inscribed angle standing on the equal arc. |
5 | PQ//SR | 5 | From statement 4 and alternate angle are equal. |
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P
S
N
M
Q
R
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Opposite angle of a parallelogram. |
2 | | 2 | Co-interior angle and PQ//QR. |
3 | | 3 | Opposite angled of a cyclic quadrilateral. |
4 | | 4 | From statements 1, 2 and 3. |
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Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Both are the right angle. |
2 | A, E, D and C are the cyclic points | 2 | Being the angles on the same segment AC are equal. |
3 | AEDC is a cyclic quadrilateral | 3 | From statements 2. |
4 | | 4 | Being the corresponding angles are equal in 3. |
A
E
B
D
C
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P
A
C
B
Q
Y
X
Proof:
S.N | Statements | S.N | reasons |
1 | | 1 | Given. |
2 | | 2 | Both are standing on the same arc. |
3 | XYQP is a cyclic quadrilateral | 3 | |
4 | | 4 | Angles at the same arc YQ. |
5 | | 5 | Angles at the same arc QB. |
6 | | 6 | From 4 and 5. |
7 | XY//AB | 7 | From 6. |
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A
P
D
B
C
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
2 | | 2 | Base angle of an isosceles triangle Or, AB=AC. |
3 | | 3 | Both are standing on the same arc. |
4 | | 4 | From statements 1, 2, and 3. |
5 | | 5 | From statement 4. |
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30.The bisector of opposite angles ∡A and ∡C of a cyclic quadrilateral ABCD intersect the corresponding circle at the points X and Y respectively. Prove that XY is the diameter of the circle.
Given: The bisector of opposite angles ∡A and ∡C of a cyclic quadrilateral ABCD intersect the corresponding circle at the points X and Y respectively.
To prove: XY is the diameter of the circle
B
C
D
A
Y
X
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Opposite arcs of equal angles. |
2 | | 2 | Same reason as 1. |
3 | | 3 | Adding the statement and 1 and 2. |
4 | | 4 | Whole part axiom. |
5 | XY is a diameter. | 5 | From 3 arcs on both the sides of PQ are equal. |
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31. In the given figure, AB and EF are parallel to each other. Prove that CDEF is a cyclic quadrilateral.
A
D
E
F
C
B
Given: AB and EF are parallel to each other
To Prove: CDEF is a cyclic quadrilateral
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
2 | | 2 | Sum of co-interior angle. |
3 | | 3 | From statements 1 and 2. |
4 | CDEF is a cyclic quadrilateral | 4 | |
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32. In the given, NPS, MAN and RMS are straight lines. Prove that PQRS is a cyclic quadrilateral.
Given: NPS, MAN and RMS are straight lines
To prove: PQRS is a cyclic quadrilateral
N
P
S
M
R
Q
A
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Inscribed angles on the same arc. |
2 | | 2 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
3 | | 3 | From statements 1 and 2. |
4 | PQRS is a cyclic quadrilateral | 4 | From statement 3. |
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33. In the given figure, AP is the radius of circle ABC and diameter of circle APQ. Prove that AQ=QC.
Q
A
P
B
C
Given: AP is the radius of circle ABC and diameter of circle APQ
To prove: AQ=QC.
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Inscribed angle of a semi circle. |
2 | AQ=QC | 2 | The perpendicular PQ from the center P upon the chord AC bisect the chord. |
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34. In the adjoining figure, X and Y are the centers of the circles which interest at A and C. XA and XC are produced to meet the other circle at B and D. Prove that AB=CD.
Given: X and Y are the centers of the circles which interest at A and C, XA and XC are produced to meet the other circle at B and D
Construction: Joined A,C and B, D
To prove: AB=CD
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Being ABCD is a cyclic quadrilateral. |
2 | | 2 | Same reason as 2. |
3 | | 3 | Radii of the same circle. |
54 | | 4 | From statements 1, 2 and 3. |
5 | XB=XD and AB=CD | 5 | Being the base angles of a triangle XBD are equal in 3. |
X
C
A
B
D
Y
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35. In the adjoining figure,, two circles intersect at A and B. The straight line PAQ meets the circles at P and Q and the straight line RBS meets the circles at R and S. Prove that PR//QS
Given: two circles intersect at A and B. The straight line PAQ meets the circles at P and Q and the straight line RBS meets the circles at R and S.
To prove: PR//QS
Construction: Join A and B.
A
B
R
P
Q
S
Proof:
S.N | Statements | S.N | Reasons |
1 | PARB is a cyclic quadrilateral | 1 | P, A, B and R are con cyclic. |
2 | ABSQ is a cyclic quadrilateral | 2 | A, B, S and Q are con cyclic. |
3 | | 3 | Exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. |
4 | | 4 | Sum of straight angles. |
5 | | 5 | From statement 3 and 4. |
6 | PR//QS | 6 | |
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36. In the given figure, two circles are intersecting at M and N. PQ and RS passes through M. If R, P, S and Q are joined to N, prove that ∡PNR=∡SNQ.
M
N
Q
S
P
R
Given: two circles are intersecting at M and N. PQ and RS passes through M. If R, P, S and Q are joined to N
To prove: ∡PNR=∡SNQ.
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Vertically opposite angles. |
2 | | 2 | Both are standing on the same arc. |
3 | | 3 | Same as reason 2. |
4 | | 4 | From statements 1, 2 and 3. |
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37. In the given figure, AP⊥BC, BR⊥AC and CQ⊥AB. Prove that ∡OPQ=∡OPR
B
Q
A
R
C
P
O
Given: Here, AP⊥BC, BR⊥AC and CQ⊥AB
To prove: ∡OPQ=∡OPR
Proof:
S.N | Statements | S.N | Reasons |
1 | PORC is a cyclic quadrilateral | 1 | |
2 | BPOQ is a cyclic quadrilateral | 2 | Same reason as 1. |
3 | B, C, R and Q are cyclic points | 3 | Being inscribed angle on BC arc equal. |
4 | | 4 | Being inscribed angle on the same arc. |
5 | | 5 | Same reason as 4. |
6 | | 6 | Same as reason 4. |
7 | | 7 | From statements 4, 5 and 6. |
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38. In the figure, O is the center of a circle. Two chords XY and MN are produced to meet at A,
then prove that ∡XOM-∡NOY=2∡MAX
A
N
Y
X
M
O
Given: O is the center of a circle. Two chords XY and MN are produced to meet at A
To prove: ∡XOM-∡NOY=2∡MAX
Construction: joined NX.
Proof:
S.N | Statements | S.N | Reasons |
1 | | 1 | Both are standing on the same arc. |
2 | | 2 | Same reason as 1. |
3 | ∡MNX=∡MAX + ∡NXY | 3 | Exterior angle of a triangle is equal to sum of its opposite two interior angles. |
4 | | 4 | From statements 1, 2 and 3. |
5 | | 5 | From statement 4 |
6 | 2∡MAX=∡XOM-∡NOY | 6 | From statement 5. |
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Q.N.39. In the given figure, O is the center of the given circle, and chords AB and CD intersect at a point E inside the circle. Prove that: ∡AOC+∡BOD=2∡AEC.
A
B
D
C
E
O
Given: O is the center of the circle, chords AB and CD are intersect at point E.
To Prove: ∡AOC+∡BOD=2∡AEC.
Construction: Joined C and B
Proof
S.N | Statements | S.N | Reasons |
1 | ∡AOC=2∡ABC | 1 | Both are standing on the same arc. |
2 | ∡BOD=2∡BCD | 2 | Same as reason 1. |
3 | ∡AOC+∡BOD=2∡BCD+2∡ABC =2(∡BCD+∡ABC) | 3 | From statements (1) and (2), Or, Adding (1) and (2) |
4 | ∡AOC+∡BOD==2∡AEC | 4 | Exterior angle of a triangle is equal to sum of its opposite two interior angles. |
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Given: O is the center of a circle and the chords AB and CD are perpendicular to each other.
Construction: Joined A and C
Proof
S.N | Statements | S.N | Reasons |
1 | | 1 | |
2 | | 2 | Sum of acute angle of rt . angled triangle. |
3 | ∡BOC=2∡MAC | 3 | Both are standing on the same arc. |
4 | ∡AOD=2∡AMC | 4 | Same reason as (3). |
5 | ∡BOC+∡AOD=2∡AMC+2∡MAC | 5 | Adding statements (3) and (4). |
6 | | 6 | From statements (4) and (5). |
O
M
D
A
C
B
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Q.N.41. In the given figure, MN is the diameter of a circle with center O, If PR=PQ, prove that ∡OSP=∡ORP.
O
M
S
R
N
Q
P
Given: O is the center and MN is the diameter of a circle.
To prove: ∡OSP=∡ORP.
Construction: Joined O and Q.
S.N | Statements | S.N | Reasons |
1 | | 1 | |
i) | OR=OQ | i) | Radii of the same circle. |
ii) | OP=OP | ii) | Common side. |
iii) | PR=PQ | iii) | From the figure (Given). |
2 | | 2 | By SSS axiom. |
3 | ∡ORP=∡OQP | 3 | |
4 | ∡OQP=∡OSP | 4 | Radii of the same circle. |
5 | ∡ORP=∡OSP | 5 | From statements (4) and (5). |
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Thank You One And All!!!