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