Carnot's Theorem |
Given: any triangle ABC Circumradius: R Inradius: r The algebraic sum of the distances from the circumcenter O to the sides: OH+OM+OE To prove: OH+OM+OE = R+r |
Proof: HN=1/2(rc-r) MD=1/2(rb-r) EF=1/2(ra-r) 2. ON+OD+OF = 3R ra+ rb+ rc - r = 4R Then: OH+OM+OE = R+r |