CIRCLE
to a circle are equal and
Radius is perpendicular to the tangent
To prove :
(i)
∠AOB
+
∠COD
=
180o
(ii)
∠AOD
+
∠BOC
=
180o
Construction :
Draw OP, OQ, OR and OS
Proof :
In ΔAPO
and
ΔASO
[Radius is perpendicular to tangent]
∠APO
=
∠ASO
=
90o
[common side]
AO
=
AO
[radius of the same circle]
OP
=
OS
∴
ΔAPO
≅
ΔASO
[RHS rule]
[c.p.c.t.]
∠AOP
=
∠AOS
A
B
C
D
P
Q
R
S
Q. Prove the opposite sides of a quadrilateral circumscribing
a circle, subtend supplementary angles at the centre of the circle.
Sum of angles is 180º
∴
O
O
P
We know, radius is perpendicular to tangent
Let us consider AB and DC
∠AOD,
∠BOC
Side AD and side BC
Side AB and side DC
Side AB subtend which
angle at centre?
∠AOB
Let us consider ΔAPO and ΔASO
Sides AD and BC subtend which angles at centre?
Let us consider
sides AD and BC
Draw OP, OQ, OR and OS
Side DC subtends which
angle at centre?
∠DOC
Whenever, we see centre and point of contact we always draw radius
Q. Prove the opposite sides of a quadrilateral circumscribing
a circle, subtend supplementary angles at the centre of the circle.
∠AOP
=
∠AOS
=
ao
Let,
a
a
b
b
c
c
d
d
A
B
C
D
P
Q
R
S
∠BOP
=
∠BOQ
=
bo
Similarly,
∠COQ
=
∠COR
=
co
∠AOB
+
∠COD
=
180º
∠AOB
+
∠BOC
+
∠COD
+
∠AOD
=
360o
[Sum of all angles at a point is 360º]
a
+
b
+
b
+
c
+
c
+
d
+
d
+
a
=
360º
∴
2a
+
2b
+
2c
=
360º
+
2d
∴
2
=
360º
(a + b + c + d)
∴
=
180º
a
+
b
+
c
+
d
…(v)
∴
(a + b)
+
=
180º
(c + d)
∴
∴
(a + d)
+
=
180º
(b + c)
[from (v)]
∠DOR
=
∠DOS
=
do
…(i)
…(ii)
…(iii)
…(iv)
∠AOD
+
∠BOC
=
180º
∴
O
a
b
b
c
c
d
a
d
a
b
c
d
a
d
b
c
It can also be written as
a + d + b + c = 180º
Sol:
To prove :
(i)
∠AOB
+
∠COD
=
180o
(ii)
∠AOD
+
∠BOC
=
180o
∴
We know, sum of all angles at a point is 360º