If AD, BE and CF are medians of △ABC, prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Question

If AD, BE and CF are medians of △ABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

KnowledgeBoat · Accepted Answer

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 = as AD is median of BC. ⇒ AB² = AP² + PD² + - 2 x x PD ⇒ AB² = AD² + - 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 = ⇒ AC² = AD² + DC² + 2PD.DC [....From (i)] ⇒ AC² = AD² + + 2 x PD x ⇒ AC² = AD² + + PD.BC ....(iii) Adding (ii) and (iii) we get, ⇒ AB² + AC² = AD² + - BC.PD + AD² + + PD.BC ⇒ AB² + AC² = 2AD² + ....(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 = as BE is median of AC. ⇒ AB² = BE² + - 2 x x QE (From v) ⇒ AB² = BE² + - 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 = as BE is median of AC. ⇒ BC² = BE² + EC² + 2QE.EC [....From (v)] ⇒ BC² = BE² + + 2 x QE x ⇒ BC² = BE² + + QE.AC ....(vii) Adding (vi) and (vii) we get, ⇒ AB² + BC² = BE² + - AC.QE + BE² + + QE.AC ⇒ AB² + BC² = 2BE² + ....(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 = as CF is median of AB. ⇒ CB² = CF² + - 2 x x FR (From ix) ⇒ CB² = CF² + - 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 = as CF is median of AB. ⇒ AC² = CF² + AF² + 2AF.FR [....From (ix)] ⇒ AC² = CF² + + 2 x FR x ⇒ AC² = CF² + + FR.AB ....(xi) Adding (x) and (xi) we get, ⇒ CB² + AC² = CF² + - AB.FR + CF² + + FR.AB ⇒ CB² + AC² = 2CF² + ....(xii) On adding (iv), (viii) and (xii) we get, AB² + AC² + AB² + BC² + CB² + AC² = 2AD² + + 2BE² + + 2CF² + ⇒ 2(AB² + BC² + CA²) = 2(AD² + BE² + CF²) + (AB² + BC² + CA²) ⇒ 2(AB² + BC² + CA²) - (AB² + BC² + CA²) = 2(AD² + BE² + CF²) ⇒ (AB² + BC² + CA²) = 2(AD² + BE² + CF²) ⇒ (AB² + BC² + CA²) = 4(AD² + BE² + CF²) Hence, proved that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Mathematics

If AD, BE and CF are medians of △ABC, prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Pythagoras Theorem

Related Questions

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

In a quadrilateral ABCD, ∠B = 90° = ∠D. Prove that
2AC² - BC² = AB² + AD² + DC².

Mathematics

If AD, BE and CF are medians of △ABC, prove that3(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2).

Pythagoras Theorem

Related Questions

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD2 = 3AB2.

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :(i) 4AD2 = 4AC2 + BC2(ii) 4BE2 = 4BC2 + AC2(iii) 4(AD2 + BE2) = 5AB2

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB2 + CD2 = AD2 + BC2.

In a quadrilateral ABCD, ∠B = 90° = ∠D. Prove that2AC2 - BC2 = AB2 + AD2 + DC2.

If AD, BE and CF are medians of △ABC, prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

In a quadrilateral ABCD, ∠B = 90° = ∠D. Prove that
2AC² - BC² = AB² + AD² + DC².