fie EM⊥AB, EP⊥DC, EN⊥CDsi EQ⊥AD
cum EF⊥(ABC), AB ⊂( ABC) si EM⊥AB⇒(T3p)EM⊥AB ⇔d(e, AB)=EM
analog si pt EN, EP, EQ
AE/AC=1/3=AM/AB (TFAin ΔACD)⇒EQ=AM(AMEQdreptunghi)=(1/3)*9=3⇒EN=9-3=6
AE/AC=1/3= (TFA in ΔACB)=EM/BC⇒EM=(1/3) *3=1⇒PE=2
deci EM=1, EP=2, EQ=3, EN=6
FE ⊥(ABC) FE ⊥orice dreapta din (ABC)
ΔFME,ΔFNE, ΔFPE,ΔFQE dreptunghice, FM, FN, FP, FQ ipotenuze si distante de la F la laturile dreptunghiului
T.Pitagora in fiecare din ac triunghiuri ne da:
FM=√5²+1²=√26
FP=√5²+2²=√29
FQ=√5²+3²=√34
FN=√5²+6²=√61
