Draw AP perpendicular to BC.
In right triangle APD,
By pythagoras theorem,
AD2 = AP2 + PD2 .......(i)
In right triangle APB,
By pythagoras theorem,
⇒ AB2 = AP2 + BP2
⇒ AB2 = AP2 + (BD - PD)2
⇒ AB2 = AP2 + BD2 + PD2 - 2BD.PD
Since, BD = $\frac{\text{BC}}{2}$ as AD is median of BC.
⇒ AB2 = AP2 + PD2 + $\Big(\frac{\text{BC}}{2}\Big)^2$ - 2 x $\frac{\text{BC}}{2}$ x PD
⇒ AB2 = AD2 + $\frac{\text{BC}^2}{4}$ - BC.PD ....(ii) (From i)
In right triangle APC,
By pythagoras theorem,
⇒ AC2 = AP2 + PC2
⇒ AC2 = AP2 + (PD + DC)2
⇒ AC2 = AP2 + PD2 + DC2 + 2PD.DC
Since, DC = $\frac{BC}{2}$
⇒ AC2 = AD2 + DC2 + 2PD.DC [....From (i)]
⇒ AC2 = AD2 + $\Big(\frac{\text{BC}}{2}\Big)^2$ + 2 x PD x $\frac{\text{BC}}{2}$
⇒ AC2 = AD2 + $\frac{\text{BC}^2}{4}$ + PD.BC ....(iii)
Adding (ii) and (iii) we get,
⇒ AB2 + AC2 = AD2 + $\frac{\text{BC}^2}{4}$ - BC.PD + AD2 + $\frac{\text{BC}^2}{4}$ + PD.BC
⇒ AB2 + AC2 = 2AD2 + $\frac{\text{BC}^2}{2}$ ....(iv)
Draw BQ perpendicular to AC,
In right triangle BQE,
By pythagoras theorem,
BE2 = BQ2 + QE2 .......(v)
In right triangle ABQ,
By pythagoras theorem,
⇒ AB2 = BQ2 + AQ2
⇒ AB2 = BQ2 + (AE - QE)2
⇒ AB2 = BQ2 + QE2 + AE2 - 2AE.QE
Since, AE = $\frac{\text{AC}}{2}$ as BE is median of AC.
⇒ AB2 = BE2 + $\Big(\frac{\text{AC}}{2}\Big)^2$ - 2 x $\frac{\text{AC}}{2}$ x QE (From v)
⇒ AB2 = BE2 + $\frac{\text{AC}^2}{4}$ - AC.QE ....(vi)
In right triangle BQC,
By pythagoras theorem,
⇒ BC2 = BQ2 + QC2
⇒ BC2 = BQ2 + (QE + EC)2
⇒ BC2 = BQ2 + QE2 + EC2 + 2QE.EC
Since, EC = $\frac{AC}{2}$ as BE is median of AC.
⇒ BC2 = BE2 + EC2 + 2QE.EC [....From (v)]
⇒ BC2 = BE2 + $\Big(\frac{\text{AC}}{2}\Big)^2$ + 2 x QE x $\frac{\text{AC}}{2}$
⇒ BC2 = BE2 + $\frac{\text{AC}^2}{4}$ + QE.AC ....(vii)
Adding (vi) and (vii) we get,
⇒ AB2 + BC2 = BE2 + $\frac{\text{AC}^2}{4}$ - AC.QE + BE2 + $\frac{\text{AC}^2}{4}$ + QE.AC
⇒ AB2 + BC2 = 2BE2 + $\frac{\text{AC}^2}{2}$ ....(viii)
Draw CR perpendicular to AB,
In right triangle CFR,
By pythagoras theorem,
CF2 = CR2 + FR2 .......(ix)
In right triangle CBR,
By pythagoras theorem,
⇒ CB2 = CR2 + RB2
⇒ CB2 = CR2 + (BF - FR)2
⇒ CB2 = CR2 + FR2 + BF2 - 2BF.FR
Since, BF = $\frac{\text{AB}}{2}$ as CF is median of AB.
⇒ CB2 = CF2 + $\Big(\frac{\text{AB}}{2}\Big)^2$ - 2 x $\frac{\text{AB}}{2}$ x FR (From ix)
⇒ CB2 = CF2 + $\frac{\text{AB}^2}{4}$ - AB.FR ....(x)
In right triangle ACR,
By pythagoras theorem,
⇒ AC2 = RC2 + AR2
⇒ AC2 = RC2 + (AF + FR)2
⇒ AC2 = RC2 + FR2 + AF2 + 2AF.FR
Since, AF = $\frac{\text{AB}}{2}$ as CF is median of AB.
⇒ AC2 = CF2 + AF2 + 2AF.FR [....From (ix)]
⇒ AC2 = CF2 + $\Big(\frac{\text{AB}}{2}\Big)^2$ + 2 x FR x $\frac{\text{AB}}{2}$
⇒ AC2 = CF2 + $\frac{\text{AB}^2}{4}$ + FR.AB ....(xi)
Adding (x) and (xi) we get,
⇒ CB2 + AC2 = CF2 + $\frac{\text{AB}^2}{4}$ - AB.FR + CF2 + $\frac{\text{AB}^2}{4}$ + FR.AB
⇒ CB2 + AC2 = 2CF2 + $\frac{\text{AB}^2}{2}$ ....(xii)
On adding (iv), (viii) and (xii) we get,
AB2 + AC2 + AB2 + BC2 + CB2 + AC2 = 2AD2 + $\frac{\text{BC}^2}{2}$ + 2BE2 + $\frac{\text{AC}^2}{2}$ + 2CF2 + $\frac{\text{AB}^2}{2}$
⇒ 2(AB2 + BC2 + CA2) = 2(AD2 + BE2 + CF2) + $\frac{1}{2}$(AB2 + BC2 + CA2)
⇒ 2(AB2 + BC2 + CA2) - $\frac{1}{2}$(AB2 + BC2 + CA2) = 2(AD2 + BE2 + CF2)
⇒ $\frac{3}{2}$(AB2 + BC2 + CA2) = 2(AD2 + BE2 + CF2)
⇒ $3$(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2)
Hence, proved that 3(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2).
