abc | ||
odpowiedź: | ||
amn+bnk+cmk |
ak+bm+an | ||
więc PΔnk=1/2nk* | ||
ac |
ak+bm+an | ||
PΔnk=1/2nk | ||
ac |
ak+bm+an | ||
PΔmk=1/2mk | ||
ab |
ak+bm+an | ||
PΔmn=1/2mn | ||
bc |
PΔABC | |
= | |
PΔnk+PΔmk+PΔmn |
nk(ak+bm+an) | mk(ak+bm+an) | mn(ak+bm+an) | ||||
[1/2(ak+bm+an)] : [1/2 | +1/2 | +1/2 | ]= | |||
ac | ab | bc |
abc | ||
=......= | ||
amn+bkn+ckm |