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Module X: Points of Tangency, MTH 606 (SU 09) 1
Part 1: Launch
Talking Points: In high school geometry courses we deal with points of tangency with respect to their
relationship with an inscribed circle. Consider the inscribed circle (or circumscribed triangle) below.
 Write three pairs of lengths that are equal in the above sketch.
 Write out the theorem that states this.
 If BF = 10, AG = 6 and the perimeter of triangle ABC is 42 units, what is the length of GC?
Module X: Points of Tangency, MTH 606 (SU 09) 2
Part 2: Exploration
In most high school geometry courses we do not spend time exploring relationships with the “excircles”
of a triangle. A triangle has three excircles. An excircle is tangent to the three lines that include the sides
of the triangle. The center of each excircle is the intersection of an angle bisector of the triangle and the
angle bisectors of the two exterior angles at the other vertices.
(1) Construct a sketch of a triangle and one of its excircles. It is important that you try this on your own
first, as it is a major focus of this module. However, if you try at length and do not get it you can go to:
http://www.geogebra.org/en/upload/files/MTH606/Lisa_James/Points%20of%20Tangency/Finding_a
n_excircle.html for assistance.
(2) Go to:
http://www.geogebra.org/en/upload/files/MTH606/Judy_Buckenmyer/PtsOfTang_Perimeter.html
to examine the sketch of triangle ABC and its three excircles. A screenshot is shown below as well.
a) Identify, once again, some lengths that are equal.
b) If the perimeter of triangle ABC were 24, what is the length of AF? If you don’t know, go back to
the geogebra sketch and measure the appropriate lengths on that triangle to help figure it out.
c) “Half the length of the perimeter” can be replaced with “semiperimeter”. Recall the standard
geometry labeling technique for sides of triangles, that is: a = BC, b = AC, and c = AB.
With this notation in mind, how could you represent BG or BF in terms of the semiperimeter
and the adjacent side?
Module X: Points of Tangency, MTH 606 (SU 09) 3
The previous questions (and their answers) lead us to the following conjectures.
Part 3: Verification
Consider the following statement.
Conjecture: The segment of a line, containing the side of a triangle, from the vertex of a triangle to the
point of contact of the opposite excircle has a length equal to half the length of the perimeter of the
triangle.
(3) Complete the steps that are missing.
The perimeter of triangle ABC = AB + BC + CA
= AB + ( _______ + ________) + CA
But BG = _______ and GC = _______, so the perimeter of triangle ABC
= AB + BF + HC + CA
But AF = HA (Why?)
= AF + HA
(4) Does the above constitute a proof? Discuss this with a colleague.
The following conjecture refers back to 2(c).
Conjecture: The segment of a side of a triangle from the vertex to the point of tangency of an excircle has
a length equal to the semiperimeter minus the length of the other adjacent side.
(5) Verify this to your own satisfaction alone or with a colleague.
Module X: Points of Tangency, MTH 606 (SU 09) 4
Part 4: Enrichment / Extension

Now let’s consider segments tangent at the points of the incircle. Go to:
http://www.geogebra.org/en/upload/files/MTH606/Lisa_James/Points%20of%20Tangency/Exc
ircles_and_incircle.html
1. Is there a relationship between the segment connecting the vertex of the triangle to point of
tangency of the incircle and the semiperimeter of the triangle? If so, define the segment in terms
of the semiperimeter.
2. Prove that the sum of the lengths of the legs of the right triangle minus the length of the
hypotenuse equals the diameter of the incircle.

The “Problem of Apollonius” is actually a series of problems in which the goal is to construct a
circle that is tangent to three objects where the objects are points, lines, and/or circles. The 10
different possibilities where P = point, L = line, and C = circle are: PPP, PPL, PPC, PLC, PLL, PCC,
LCC, LLC, LLL, and CCC. Some scenarios have multiple cases.
By constructing the incircle and excircles today we have performed the LLL task with three
intersecting lines. Now let’s employ the WIN strategy and consider the LLL construction if the 3
lines are not intersecting. Consider the drawing below. What would your strategy be to construct
a circle tangent to the 3 lines if 2 lines are parallel? What would happen if all 3 lines were parallel?
Try it on your own first, but if you want to see a solution to the problem with 2 parallel and 1
intersecting line, go to the link below the diagram. For additional “Problem of Apollonius”
constructions see chapter 10 in the Posamentier text.
http://www.geogebra.org/en/upload/files/MTH606/Judy_Buckenmyer/PtsOfTangency_Apollonius.html