Răspuns:
Explicație pas cu pas:
in ΔABC
m(∡B)=90°
aplicam teorema lui Pitagora
AB²+BC²=AC²
40²+30²=AC²
1600+900=AC²
2500=AC²=>AC=√2500=>AC=50
in ΔABC, BE⊥AC
aplicam teorema catetei
AD²=AF × AC
30²= AF×50
900=AE×50
AF=900:50
AE=18
in ΔADC, m(∡D)=90° si DF⊥AC
aplicam teorema catetei
BC²=EC×AC
30²=EC× 50
900=EC× 50
EC=900:50
EC=18
AC=AE+EF+EC
50=18+EF+18
50=36+EF
EF=50-36
EF=14
