Circumscribed and Inscribed Circles
•
Section 2.5
63
We will use Figure 2.5.6 to find the radius
r
of the inscribed circle. Since
OA
bisects
A
,
we see that tan
1
2
A
=
r
AD
, and so
r
=
AD
·
tan
1
2
A
. Now,
△
OAD
and
△
OAF
are equivalent
triangles, so
AD
=
AF
. Similarly,
DB
=
EB
and
FC
=
CE
. Thus, if we let
s
=
1
2
(
a
+
b
+
c
), we
see that
2
s
=
a
+
b
+
c
=
(
AD
+
DB
)
+
(
CE
+
EB
)
+
(
AF
+
FC
)
=
AD
+
EB
+
CE
+
EB
+
AD
+
CE
=
2(
AD
+
EB
+
CE
)
s
=
AD
+
EB
+
CE
=
AD
+
a
AD
=
s
−
a
.
Hence,
r
=
(
s
−
a
) tan
1
2
A
. Similar arguments for the angles
B
and
C
give us:
Theorem 2.10.
For any triangle
△
ABC
, let
s
=
1
2
(
a
+
b
+
c
). Then the radius
r
of its inscribed
circle is
r
=
(
s
−
a
) tan
1
2
A
=
(
s
−
b
) tan
1
2
B
=
(
s
−
c
) tan
1
2
C
.
(2.38)
We also see from Figure 2.5.6 that the area of the triangle
△
AOB
is
Area(
△
AOB
)
=
1
2
base
×
height
=
1
2
c r
.
Similarly, Area(
△
BOC
)
=
1
2
ar
and Area(
△
AOC
)
=
1
2
b r
. Thus, the area
K
of
△
ABC
is
K
=
Area(
△
AOB
)
+
Area(
△
BOC
)
+
Area(
△
AOC
)
=
1
2
c r
+
1
2
ar
+
1
2
b r
=
1
2
(
a
+
b
+
c
)
r
=
sr
,
so by Heron’s formula we get
r
=
K
s
=
radicallow
s
(
s
−
a
)(
s
−
b
)(
s
−
c
)
s
=
radicalBigg
s
(
s
−
a
)(
s
−
b
)(
s
−
c
)
s
2
=
radicalBigg
(
s
−
a
)(
s
−
b
)(
s
−
c
)
s
.
We have thus proved the following theorem:
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 Fall '11
 Dr.Cheun
 Calculus, Geometry, Angles, AD + EB

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