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5. Relationships in Triangles
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United States of AmericaFL
Grade 912

5.04 Centers of triangles

Lesson
Practice
Lesson

Concept summary

There are two special circles that can be defined in relation to triangles.

Circumscribed circle

The smallest circle that includes a plane figure. If the figure is a polygon, then the circle must contain all of the vertices of the polygon.

A triangle inside a circle. Each vertex of the triangle lies on the circle
Inscribed circle

The largest possible circle that can be contained in a plane figure. If the plane figure is a polygon, then the circle must be tangent to all of the sides of the polygon.

A circle inside of a triangle. Each side of the triangle intersects the circle at exactly one point

A point of concurrency is a point where three or more lines coincide. We can define different centers of triangles as the point of concurrency of different lines and line segments in triangles.

Incenter

The point of concurrency of the angle bisectors of a triangle. It is called the incenter because it is the center of the inscribed circle of the triangle.

A triangle with a dashed line segment coming from each angle and intersecting the opposite side. Each segment cuts the angle perfectly in half and the 3 segments intersect at a single point
Incenter theorem

The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.

Circumcenter

The point of concurrency of the perpendicular bisectors of the triangle. It is called the circumcenter because it is the center of the circumscribed circle of the triangle.

A triangle where each side has a line passing through it. Each line splits the side it passes through exactly in half and intersects the side at a 90 degree angle. All 3 lines intersect at a single point
Circumcenter theorem

The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.

Centroid

The point of concurrency of the three medians of a triangle.

A triangle with a dashed line segment coming from each angle and intersecting the opposite side at the point that cuts the side perfectly in half. The 3 segments intersect at one point
Centroid theorem

The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.

The centers do not always lie inside the triangle - it depends on the type of triangle.

For an equilateral triangle, the incenter, circumcenter, and centroid all coincide at the same point.

Worked examples

Example 1

G is the centroid of the triangle.

a

Describe the relationship between \overline{CD} and \overline{BD}.

Solution

We are given that G is the centroid, which is the intersection of the medians of the triangle. So D must be the midpoint of \overline{BC}.

So then, \overline{CD} \cong \overline{BD}.

b

Find the value of x.

Approach

We now know that \overline{CD} \cong \overline{BD}, so we can set their lengths to be equal and solve the equation for x.

Solution

\displaystyle CD\displaystyle =\displaystyle BD\overline{CD} \cong \overline{BD}
\displaystyle 4x+5\displaystyle =\displaystyle x+11Substitute in the length of the segements
\displaystyle 3x\displaystyle =\displaystyle 6Subtract x and 5 from both sides
\displaystyle x\displaystyle =\displaystyle 2Divide both sides by 3
c

Find the length of \overline{BG}.

Approach

The centroid theorem tells us that BG=\dfrac{2}{3}BE which means that EG:GB=1:2. We also know that BE=BG+GE by the segment addition postulate. We can calculate GE and then use that to calculate BG.

Solution

Substitute x=2 into the equation GE=3x+1 to find that GE=7.

We can then compute that

BG=2GE=2\cdot 7which we can simplify to find BG=14.

Reflection

Since we know the ratio of the lengths of \overline{BG}, \overline{GE} and \overline{BE}, knowing any one of these lengths is enough to find the other two.

Example 2

P is the incenter of the triangle.

a

Determine m \angle BAP.

Approach

Since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. So we know that \overline{AP} bisects \angle BAC, and therefore m\angle BAP = \dfrac{1}{2} m\angle BAC.

We can use the measures of the two known angles to find the measure of \angle BAC.

Solution

\displaystyle m\angle BAC + m\angle ABC + m\angle ACB\displaystyle =\displaystyle 180\degreeSum of internal angle measures of a triangle
\displaystyle m\angle BAC + 76\degree + 62\degree\displaystyle =\displaystyle 180\degreeSubstitute known angle measures
\displaystyle m\angle BAC\displaystyle =\displaystyle 42\degreeSubtract 138\degree from both sides

Then we have

\displaystyle m\angle BAP\displaystyle =\displaystyle \frac{1}{2} m\angle BAC\overline{AP} bisects \angle BAC
\displaystyle m\angle BAP\displaystyle =\displaystyle \frac{1}{2} (42\degree)Substitution
\displaystyle m\angle BAP\displaystyle =\displaystyle 21 \degreeSimplify
b

Determine m \angle BPC.

Approach

Again, since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. We can use this to find the measures of \angle PBC and \angle PCB, and then use those to determine the measure of \angle BPC.

Solution

\displaystyle m\angle PBC\displaystyle =\displaystyle \frac{1}{2} m\angle ABC\overline{PB} bisects \angle ABC
\displaystyle =\displaystyle \frac{1}{2} \left(76 \degree\right)Substitute angle measure
\displaystyle =\displaystyle 38 \degreeSimplify expression

and

\displaystyle m\angle PCB\displaystyle =\displaystyle \frac{1}{2} m\angle ACB\overline{PC} bisects \angle ACB
\displaystyle =\displaystyle \frac{1}{2} \left(62 \degree\right)Substitute angle measure
\displaystyle =\displaystyle 31 \degreeSimplify expression

Then we have

\displaystyle m\angle BPC + m\angle PBC + m\angle PCB\displaystyle =\displaystyle 180\degreeSum of internal angle measures of a triangle
\displaystyle m\angle BPC + 38\degree + 31\degree\displaystyle =\displaystyle 180\degreeSubstitute known angle measures
\displaystyle m\angle BPC\displaystyle =\displaystyle 111\degreeSubtract 69\degree from both sides

Outcomes

MA.912.GR.5.2

Construct the bisector of a segment or an angle, including the perpendicular bisector of a line segment.

MA.912.GR.5.3

Construct the inscribed and circumscribed circles of a triangle.

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