1
Circumcircles and Incircles of Triangles
I. Circumcircle of a Triangle
Theorem: All triangles are cyclic,
i.e. every triangle has a circumscribed
circle or circumcircle.
Proof
Given ABC, construct the perpendicular bisectors of
sides AB and AC.
(Recall that a perpendicular bisector is a line that
forms a right angle with one of the triangle's sides and
intersects that side at its midpoint.)
These bisectors will intersect at a point O.
So we have that OD t AB and OE t AC.
Now observe that
ODA y
Thus,
OA = OB
It is also the case that
OEA y
So
ODB by the Side-Angle-Side Theorem.
being corresponding sides of congruent triangles.
OEC by the Side-Angle-Side Theorem.
OA = OC being corresponding sides of congruent triangles.
Now consider any point X on segment AE. We find from the Pythagorean Theorem that
OA =
OE 2 + EA2 =
OE 2 + ( EX + XA ) 2
>
OE 2 + EX 2 = OX .
Therefore, OA > OX.
In a similar way, we can establish that r = OA = OB = OC is greater than the distance from O
to any other point Z on ABC.
Hence, the circle with center at O and radius r circumscribes the triangle.
2
The cirrcumcenterr's position depends
d
on the type of triangle
i. If and
a only if a triangle is acute
a
(alll angles smaaller than a right
r
angle),
the circumcenteer lies insidee the trianglee.
ii. If and
a only if itt is obtuse
(ha
as one angle bigger than a right anglle),
the circumcenteer lies outsidde the trianglle.
iii. Iff and only if it is a right triangle,
the circumcennter lies at thhe
ceenter of the hypotenuse.
h
Theorem
m: The ratio that appearss in the Law of Sines is the diameter of the circum
mcircle of
ABC:
d =
c
a
b
=
=
sin α
sin β
ssin γ
Proof
Given
ABC , let O denote thhe center of its
circumcirrcle. Observ
ve that :BO
OC = 2 × :BAC
:
due to the Inscribed and
a Central Angle Theorem.
That is,
:B
BOC = 2 α . (1)
Let M be
b the midpo
oint of BC.
Then BOM
B
y
Theorem.
COM
C
by the Side-Side-Siide
α
O
α
3
So we have that
:BOM =:COM =
1
2
:BOC , (2)
since corresponding angles in congruent triangles are equal.
For the same reason, we know that
:OMB =:OMC =
1
2
:BMC =
1
ο
× 180
2
= 90ο .
From (1) and (2) it follows immediately that :BOM = α .
Now in the right triangle
sin α =
Therefore,
d =
BOM we see that
2 ⋅ BM
a
BM
BM + MC
BC
= .
=
=
=
2 ⋅ BO
2⋅r
2⋅r
BO
d
a
.
sin α
Hence, by applying the Law of Sines, we may conclude that
d =
a
b
c
.
=
=
sin α
sin β
sin γ
II. Incircle of a Triangle
Theorem: A circle can be inscribed in any triangle,
i.e. every triangle has an incircle.
Proof
Given
ABC , bisect the angles at the vertices
A and B. These angle bisectors must intersect at a
point, O . Locate the points D , E and F on sides
AB, BC and CA respectively so that
OD t AB, OE t BC and OF t CA.
Observe that
AOD y
AOF
and
BOD y
BOE
by the Angle-Side-Angle Theorem.
Since corresponding sides of congruent triangles are equal, we also know that
OD = OF
and
OD = OE.
4
Hence,
OD = OF = OE .
Moreover, it follows from the Pythagorean Theorem, that r = OD = OF = OE is the shortest
distance from the point O to each of the sides of ABC.
So the circle with center O and radius r is an incircle for the triangle. Further, it is the only
one, since any point equidistant from segments AB and BC must necessarily lie on line OB: the
bisector of :ABC ; and similarly, any point equidistant from segments CA and AB must lie
on line OA: the bisector of :CAB . Therefore, O must be the center of the incircle for the
ABC.
Consequently, the incircle for any triangle is unique.
Theorem: The radius r of the incircle for
r =
ABC is given by
2 × Area ( ABC )
Perimeter ( ABC )
.
Proof
Given
ABC with incircle having center at O, we note that
Area ( ABC ) = Area ( AOB ) + Area ( BOC ) + Area ( COA ) .
Thus,
1
2
( AB × r ) +
Area ( ABC ) =
r
2
( AB + BC + CA ) .
Area ( ABC ) =
r
× Perimeter (
2
Area ( ABC ) =
1
2
( BC × r ) +
That is,
So,
Therefore,
r =
ABC ).
2 × Area ( ABC )
Perimeter ( ABC )
.
1
2
( CA × r ) .
5
Exercises
1. Find the radii of each of the three circles in the figure below.
2. Find the radii of each of the circles in the given equilateral triangle. Take the side length to be
1 unit in each case.
a.
b.