1
Circumcircles and Incircles of Triangles Unlock Access to An
Exclusive 30 Theorem : All triangles are cyclic, Day Trial i.e. every triangle has a circumscribed I. Circumcircle of a Triangle
C&I_Tri
circle or circumcircle .
Proof Access Now Given ABC , construct the perpendicular bisectors of sides AB and AC . No thanks, I don't want my exclusive trial ( Recall 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 ODB by the Side-Angle-Side Theorem. Thus,
OA = OB
being corresponding sides of congruent triangles.
It is also the case that OEA y OEC by the Side-Angle-Side Theorem. So
OA = OC being corresponding sides of congruent triangles.
Now consider any point X on on segment AE . We find from the Pythagorean Theorem that OA =
Therefore,
OE 2
+
EA2
=
2
OE
+
( EX + XA ) 2 >
OE 2
+
EX 2
= OX . .
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 ci cumcenter's position epends on the type of triangle
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i. If and only if a triangle is acute (al angles smaller than a ight angle), the circumcent r lies insid the triangl .
Exclusive 30 Day Trial C&I_Tri
Access Now
ii. If nd only if it is obtuse No thanks, I don't wantang my exclusive trial (has one angle bigger than a right e), the circumcent r lies outsi e the triang e.
iii. If and only if it is a right triangle, the circumcenter lies at the center of the ypotenuse.
Theore : The ratio that appears in the Law of Sines is the diameter of the circu circle of ABC : d
=
a sin α
=
b sin β
=
c in γ
Proof Given ABC , let O denote t e center of its circumcircle. Obser e that : B C = 2 × BAC due to the Inscribed nd Central Angle Theorem. That is, : BOC = 2 .
Let M e the midpoint of BC . Then OM y OM by the Side-Side-S de Theorem.
O
3 So we have that : BOM
1
=:COM =
: BOC ,
2
Access to An since corresponding angles in congruentUnlock triangles are equal.
Exclusive 30 BMC = 0 90 Day Trial
For the same reason, we know that :OMB
=:OMC =
1 2
1
:
×
sin α
Therefore,
d
=
=
BM BO
BOM
=
ο
ο
=
.
2
From (1) and (2) it follows immediately that Now in the right triangle
18
: BOM C&I_Tri
= α .
we see that Access Now
2 ⋅ BM BM + MC BC a . = = = 2 ⋅ BONo thanks, 2 ⋅I r 2 ⋅ r d don't want my exclusive trial
a . sinα
Hence, by applying the Law of Sines, we may conclude that d
=
a sin α
=
b sin β
=
c . 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 .
Unlock Accessthat to Anr = OD = OF = OE is the shortest Moreover, it follows from the Pythagorean Theorem, distance from the point O to each of the sides of ABC.
Exclusive 30 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 Day Trial bisector of ABC ; and similarly , any point equidistant from segments CA and AB must lie :
on line OA: the bisector of :CAB . Therefore,C&I_Tri O must be the center of the incircle for the ABC .
Now Consequently, the incircle for any triangle isAccess unique. No thanks, I don't want my exclusive trial
Theorem : The radius r of the incircle for r =
ABC
2 × Area ( ABC ) Perimeter ( ABC )
is given by
.
Proof
Given
ABC
with incircle having center at O, we note that
Area ( ABC ) = Area ( AOB ) + Area ( BOC ) + Area (COA ) .
Thus, Area ( ABC ) =
1 2
( AB
×
r) +
1 2
( BC
×
r) +
That is, Area ( ABC ) =
r
2
( AB + BC + CA ) .
So, Area ( ABC ) =
r
2
×
Perimeter ( ABC ).
Therefore, r =
2 × Area ( ABC ) Perimeter ( ABC )
.
1 2
( CA
×
r).
5
Exercises
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Exclusive 30 Day Trial
1. Find the radii of each of the three circles in the figure below.
C&I_Tri
Access Now No thanks, I don't want my exclusive trial
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.
