If AD, BE and CF are medians of △ABC, prove that

3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

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 = $\dfrac{\text{BC}}{2}$ as AD is median of BC.

⇒ AB² = AP² + PD² + $\Big(\dfrac{\text{BC}}{2}\Big)^2$ - 2 x $\dfrac{\text{BC}}{2}$ x PD

⇒ AB² = AD² + $\dfrac{\text{BC}^2}{4}$ - 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 = $\dfrac{BC}{2}$

⇒ AC² = AD² + DC² + 2PD.DC [….From (i)]

⇒ AC² = AD² + $\Big(\dfrac{\text{BC}}{2}\Big)^2$ + 2 x PD x $\dfrac{\text{BC}}{2}$

⇒ AC² = AD² + $\dfrac{\text{BC}^2}{4}$ + PD.BC ….(iii)

Adding (ii) and (iii) we get,

⇒ AB² + AC² = AD² + $\dfrac{\text{BC}^2}{4}$ - BC.PD + AD² + $\dfrac{\text{BC}^2}{4}$ + PD.BC

⇒ AB² + AC² = 2AD² + $\dfrac{\text{BC}^2}{2}$ ….(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 = $\dfrac{\text{AC}}{2}$ as BE is median of AC.

⇒ AB² = BE² + $\Big(\dfrac{\text{AC}}{2}\Big)^2$ - 2 x $\dfrac{\text{AC}}{2}$ x QE (From v)

⇒ AB² = BE² + $\dfrac{\text{AC}^2}{4}$ - 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 = $\dfrac{AC}{2}$ as BE is median of AC.

⇒ BC² = BE² + EC² + 2QE.EC [….From (v)]

⇒ BC² = BE² + $\Big(\dfrac{\text{AC}}{2}\Big)^2$ + 2 x QE x $\dfrac{\text{AC}}{2}$

⇒ BC² = BE² + $\dfrac{\text{AC}^2}{4}$ + QE.AC ….(vii)

Adding (vi) and (vii) we get,

⇒ AB² + BC² = BE² + $\dfrac{\text{AC}^2}{4}$ - AC.QE + BE² + $\dfrac{\text{AC}^2}{4}$ + QE.AC

⇒ AB² + BC² = 2BE² + $\dfrac{\text{AC}^2}{2}$ ….(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 = $\dfrac{\text{AB}}{2}$ as CF is median of AB.

⇒ CB² = CF² + $\Big(\dfrac{\text{AB}}{2}\Big)^2$ - 2 x $\dfrac{\text{AB}}{2}$ x FR (From ix)

⇒ CB² = CF² + $\dfrac{\text{AB}^2}{4}$ - 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 = $\dfrac{\text{AB}}{2}$ as CF is median of AB.

⇒ AC² = CF² + AF² + 2AF.FR [….From (ix)]

⇒ AC² = CF² + $\Big(\dfrac{\text{AB}}{2}\Big)^2$ + 2 x FR x $\dfrac{\text{AB}}{2}$

⇒ AC² = CF² + $\dfrac{\text{AB}^2}{4}$ + FR.AB ….(xi)

Adding (x) and (xi) we get,

⇒ CB² + AC² = CF² + $\dfrac{\text{AB}^2}{4}$ - AB.FR + CF² + $\dfrac{\text{AB}^2}{4}$ + FR.AB

⇒ CB² + AC² = 2CF² + $\dfrac{\text{AB}^2}{2}$ ….(xii)

On adding (iv), (viii) and (xii) we get,

AB² + AC² + AB² + BC² + CB² + AC² = 2AD² + $\dfrac{\text{BC}^2}{2}$ + 2BE² + $\dfrac{\text{AC}^2}{2}$ + 2CF² + $\dfrac{\text{AB}^2}{2}$

⇒ 2(AB² + BC² + CA²) = 2(AD² + BE² + CF²) + $\dfrac{1}{2}$(AB² + BC² + CA²)

⇒ 2(AB² + BC² + CA²) - $\dfrac{1}{2}$(AB² + BC² + CA²) = 2(AD² + BE² + CF²)

⇒ $\dfrac{3}{2}$(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).