In tr. GDF
Ai [GC] mediana
Deci, A[GDC]=A[GCF]
Tr. GDC e dr. isosc. =>A[GDC]=(a·a)/2
Deci, A[GCF]=(a·a)/2
Triunghiurile GCF, GDH, EAH si EBF sunt congruente
=> A[GCF]=A[GDH]=A[EAH]=A[EBF]=(a·a)/2
Si
A[GCF]+A[GDH]+A[EAH]+A[EBF]=4·(a·a)/2=2·a·a=2·80·80=2·6400=12800(m^2)