[size=150][size=100]This line intersects the circumcircle at D and E, and the incircle at F and G. [br]r[sup]2[/sup] = DF x EG[/size][br][/size][sup][/sup]
[size=150][code][/code]A line though the midpoint D and the incenter I intersects the altitude from C at F. The distance from F to C is equal to the radius of the incircle[br][/size][sup][/sup]
[size=150]The incenter I cuts the angle bisector CD into two parts where[br]DI : IC = AB : AC + CB [br][/size][sup][/sup]
[size=150]Let points E, F, and G be the excenters of triangle ABC and I the incenter. The sum of the squares of the distances from these centers to the orthocenter is triple the square of the diameter of the circumcircle.[br]OI[sup]2[/sup] + OD[sup]2[/sup] + OE[sup]2[/sup] + OF[sup]2[/sup] = 3 x (Circumcircle Diameter)[sup]2[br][/sup][/size][sup][/sup]
[size=150]The product of the distances from the incenter to the triangle vertices is equal to the product of the square of the incircle diameter and the circumcircle radius[br][br]IA x IB x IC = d[sup]2[/sup] x R[br][/size]
[size=150]This is Euler's Triangle Formula[br][br]The distance from the incenter to the circumcenter is given by [br][math]\sqrt{R\left(R-2r\right)}[/math] where R is the circumradius and r is the inradius.[br][br][/size]
[size=150]The area of a triangle can be found by multiplying the inradius by the semiperimeter of the triangle.[/size]