Length of Tangent to Excircle

Theorem

The excircle on one side of a triangle is tangent to that side and divides it into the same two lengths as the tangent to the incircle, but in opposite order.

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Proof

Let the sides opposite vertex $A, B, C$ be $a,b,c$.

Let the semiperimeter of $\triangle ABC$ be $s$.

Construct the incircle of $\triangle ABC$ with center $M$.

Let $\triangle ABC$ be tangent to the incircle at $H$.

Construct the excircle of $\triangle ABC$ on side $c$ with center $N$.

Let $\triangle ABC$ be tangent to the excircle at $T$.

By Tangent Points of Incircle in Terms of Semiperimeter:

$AF = AH = s - a$

$CF = CG = s - c$

$BG = BH = s - b$

By addition:

$AB = AH + BH = \paren {s - a} + \paren {s - b}$

$AB = AT + BT = \paren {s - a} + \paren {s - b}$

By Tangents to Circle from Point are of Equal Length:

$AD = AT$

$BE = BT$

By addition:

$AT + BT = c$

Substituting:

$AD + BE = c$

Rearranging:

$BE = c - AD$

By Tangents to Circle from Point are of Equal Length, the tangents drawn to the excircle from $C$ are equal:

$AD + b = BE + a$

By addition:

$AD + b = c - AD + a$

$2AD = a + c - b$

$AD = \dfrac {a + c - b} 2$

By the definition of $s$ above:

$\paren {s - b} = \dfrac {a + c + b} 2 - b = \dfrac {a + c - b} 2$

Combining results:

$AD = s - b$

We find one pair of tangents to the incircle by substitution:

$AT = AD = s - b$

Since $c = \paren {s - a} + \paren {s - b} = AD + BE$:

$BE = s - a$

We find a second pair of tangents to the incircle by substitution:

$BT = BE = s - a$

The third pair of tangents $CD$ and $CE$ are

$CD = CE = 3 s - \paren {a + b + c} = s$

$\blacksquare$