?$證明:連接EF，設PF=m，PE=n.$?
?$∵點P是△ABC的重心,$?
?$∴E、F分別為AC、BC的中點$?
?$∴EF為△ABC的中位線.$?
?$∴EF∥/AB,EF=\frac{1}{2}AB,AE=\frac{1}{2}AC,BF=\frac{1}{2}BC.$?
?$∵BC=a,AC=b,AB=C,$?
?$∴AE=\frac{1}{2}b,BF=\frac{1}{2}a,EF=\frac{1}{2}\ \mathrm {c}.$?
?$∵EF∥AB，$?
?$∴∠EFP=∠BAP,∠FEP=∠ABP.\ $?
?$∴△EFP∽△BAP.$?
?$∴\frac{PE}{PB}=\frac{PF}{PA}=\frac{EF}{BA}=\frac{1}{2}，即\frac{n}{PB}=\frac{m}{PA}=\frac{1}{2}$?
?$∴PB=2n，PA=2m.$?
?$∵ AF⊥BE,$?
?$∴在Rt△AEP中，由勾股定理，得PE2+PA2=AE2,得n2+4m2=\frac{1}{4}b2①;$?
?$在Rt△BFP中，由勾股定理，得PF2+PB2=BF2,得m2+4n2=\frac{1}{4}a2②.$?
?$①+②,得5(n2+m2)=\frac{1}{4}(a2+b2).$?
?$在Rt△EFP中,由勾股定理，得PE2+PF2=EF2,得n2+m2=\frac{1}{4}c2.$?
?$∴ 5×\frac{1}{4}c2=\frac{1}{4}(a2+b2)，$?
?$即a2+b2=5c2$?
