teorem inaltimii AD
AD^2=12 x 24
AD=12√2
aria tr. At
At=BC x AD/2 = (36 x 12√2)/2
At=216√2
teorema catetei AB
AB^2=BD x BC = 12 x 36
AB=12√3
teorema catetei AC
AC^2=DC x BC=24 x 36
AC=12√6
mediana BE, E∈AC, cu pitagora in BAE
BE=√(AB^2+(AC/2)^2) = √(12 x 12 x 3+ 36 x 6)
BE=18√2
bisectoarea CF , F∈AB
notam x=AF, y=FB
teorema bisectoarei CF
x/y=AC/BC
x+y=AB
x/y=√6 /3
x+y=12√3
y=12√3 - x
x=y(√6 /3)
x=(12√3 - x) x √6/3
x=12√2(3-√6)
bisectoarea FC
FC=√(AC^2+x^2)=√(12 x 12 x 6+ 12 x 12 x 2 (15-6√6))
FC=24√(9-3√6), forma asta poate fi scrisa altfel daca stii sa descompui radicalii compusi
verifica si tu mai ales la calculul bisectoarei unde sincer m-am plictisit
