A circle inside triangle ABC touches AB at F, BC at D and CA at E…

When a circle touches the three sides of a triangle, the tangents from one corner are equal: $AF = AE$, $BF = BD$ and $CD = CE$. For a right triangle with the right angle at B, the radius of the circle inside is $r = \dfrac{AB + BC - AC}{2}$. When a circle touches the four sides of ABCD, then $AB + CD = BC + DA$. For two circles with the same centre, a chord of the larger that touches the smaller has half-length $\sqrt{R^2 - r^2}$.