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The equal-tangents property

From any external point, the two tangent segments drawn to a circle are always equal in length. For a triangle's incircle, this fact alone is enough to set up a full system: label the three tangent lengths from each vertex as $x, y, z$ , and the sides become $x+y$ , $y+z$ , and $z+x$ .

Incircle vs. excircles

The incircle sits inside the triangle, tangent to all three sides. Each excircle sits outside the triangle, tangent to one side and to the extensions of the other two — there is exactly one excircle per vertex (opposite that vertex), so three in total.

r = \dfrac{\text{Area}}{s} \qquad\text{(inradius)}, \qquad r_a = \dfrac{\text{Area}}{s-a} \qquad\text{(exradius opposite vertex } A\text{)}

where $s$ is the semi-perimeter. Also useful: $\text{Area} = r \cdot s$ , directly from the inradius formula rearranged.

Worked example

Problem: A triangle has sides $AB=7$ , $BC=8$ , $CA=9$ . The incircle touches side $AB$ at point $P$ . Find $AP$ .

Solution: Let the tangent lengths from $A, B, C$ be $x, y, z$ respectively. Then:

x+y = AB = 7, \qquad y+z=BC=8, \qquad z+x=CA=9

Adding all three: $2(x+y+z) = 24 \Rightarrow x+y+z=12$ . Subtracting the equation $y+z=8$ :

x = 12 - 8 = 4

Since $AP$ is the tangent length from $A$ , $AP = x = 4$ .

Practice problems

1.A triangle has sides 6, 8, 10 and area 24. Find its inradius.

2.Using the same triangle (sides 6, 8, 10, area 24), find the exradius opposite the side of length 10.

3.A triangle has sides 5, 6, 7. The incircle touches the side of length 6 at point $P$ , splitting it into two tangent-length pieces. If the tangent length from the vertex on the side of length 5 is $x$ , set up (but don't necessarily solve numerically) the system that finds it — then solve it.

4.Why must the two tangent segments from an external point to a circle always be equal?

Tangency, Incircles & Excircles

The equal-tangents property

Incircle vs. excircles

Practice problems