Draw lines connecting vertices A, B, C to O, the centre of the incircle. This divides triangle ABC into three triangles: AOB, BOC, COA, with areas given by:
2 K(AOB) = r AB
2 K(BOC) = r BC
2 K(COA) = r CA
where r is the radius of the incircle. So
(Eq 1)
2 K(ABC) = 2 K(AOB) + 2 K(BOC) + 2 K(COA) = r (AB + BC + CA)
We also have
2 K(ABC) = h_A BC = h_B CA = h_C AB
where h_X is the altitude originating from vertex X= A, B, or C. So
BC = 2 K(ABC)/h_A, CA = 2 K(ABC)/h_B, AB = 2 K(ABC) /h_C
Substituting into Eq 1:
r ( 2 K(ABC)/h_C + 2 K(ABC)/h_B + 2 K(ABC)/h_A ) = 2 K(ABC)
and
1/r = 1/h_A + 1/h_B + 1/h_C
Edited on May 9, 2020, 8:49 pm
|
|
Posted by FrankM
on 2020-05-09 20:47:46 |
Solution with explanation
