Tangent Points of Incircle in Terms of Semiperimeter
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Theorem
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $s$ denote the semiperimeter of $\triangle ABC$.
Then the distance from $A$ to a point tangent to the incircle is equal to $s - a$.
Proof
Let $D$, $E$ and $F$ be the points where the incircle is tangent to the sides $AC$, $AB$ and $BC$ respectively.
Then:
| \(\ds AD\) | \(=\) | \(\ds AE\) | Tangents to Circle from Point are of Equal Length | |||||||||||
| \(\ds CD\) | \(=\) | \(\ds CF\) | Tangents to Circle from Point are of Equal Length | |||||||||||
| \(\ds BE\) | \(=\) | \(\ds BF\) | Tangents to Circle from Point are of Equal Length | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds BF + FC + AD\) | \(=\) | \(\ds s\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a + AD\) | \(=\) | \(\ds s\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds AD\) | \(=\) | \(\ds s - a\) |
The above argument can be used mutatis mutandis to show that:
| \(\ds BE\) | \(=\) | \(\ds s - b\) | ||||||||||||
| \(\ds CF\) | \(=\) | \(\ds s - c\) |
$\blacksquare$