内切三角形和边长计算

Here I'm going to explain how to find the lengths to the tangency points to the incircle in a triangle $ABC$. Here's a diagram for reference:

More specifically, we will find the lengths $AE, EC, CD, DB, BF, FA$. Let the lengths of $BC, AC, AB$ be $a, b, c$ (just like how we do it normally).

As you see in the diagram, the incircle is tangent to all three sides of the triangle.

Similarly, $AE = AF$ and $CD = CF$. Let $AE = AF = x, BD = BF = y, CD = CE = z$. We're now going to find a way to solve for $x, y, z$.

So now we have 3 variables ($x, y, z$) but now we have three equations: $a = y+z, b = x+z, c = x+y$. This means that we can solve for $x, y, z$.

Here's a cool way to solve the system of equations above. First add all three equations together. This gives us $a+b+c = 2(x+y+z) \implies x+y+z = \frac{a+b+c}{2}$. But $y+z = a$!. So $x = \frac{a+b+c}{2} - (y+z) = \frac{b+c-a}{2}$.

Similarly, $y = \frac{a+c-b}{2}, z = \frac{a+b-c}{2}$. In fact, if we define $s$ to be the semiperimeter (half the perimeter), then $s = \frac{a+b+c}{2}$. This way, we can actually rewrite $x, y, z$ as $x = s-a, y = s-b, z = s-c$.

So this is how you find the lengths in an incircle. :)

Here's a related exercise: https://www.expii.com/exercise/1882/