in ΔABC aplic teorema lui Pitagora
AB²+BC²=AC²
4²+6²=AC²
16+36=AC²
52=AC²=>AC=√52=>AC=2√13
AM+MC=AC
CM/AM=1/2=>AM=2CM
2CM+CM=AC
3CM=AC⇔3CM=2√13=>CM=2√13/3
AM+CM=AC
AM+AM/2=AC
2AM/2+AM/2=AC
3AM/2=AC⇔3AM/2=2√13=>AM=4√14/3
duc MZ⊥BC cu Z∈BC.
AC/MC=BC/ZC
2√13/2√13/3=6/ZC
2√13:2√13/3=6/ZC
2√13×3/2√13=6/ZC
3=6/ZC=>ZC=6/3=>ZC=2
in ΔMZC: MZ²+ZC²=MC²
MZ²+2²=(2√13/3)²
MZ²+4=52/9=> MZ²=52/9-4=> MZ²=52/9-4×9/9=> MZ²=52-36/9=> MZ²=16/6=> MZ=√16/9=> MZ=4/3
aria ΔBMC=BC×MZ/2=(6×4/3)/2=8/2=4
Daca BC=BZ+ZC=> 6= BZ+2=> BZ=6-2=> BZ=4
ZC║MU, unde U∈DC si MU⊥DC, ZC=MU=>MU=2
aria ΔMDC=DC×MU/2=4×42/2=4
daca nu te ajuta sterge rasp
