Question

prove it 2(AD2+BE2+CF2)=3BC2​

Answer 1

Solution:

Given: In Δ ABC , AD , BE and CF are medians. Also, ∠A=90°

To prove: 2(AD²+BE²+CF²)=3 BC²

Construction: Draw, AM ⊥BC, BN ⊥CA,and CP⊥ AB.

Proof: In right Δ AMD, ΔAMC, and Δ AMB, and ΔABC, using Pythagoras theorem

1. AD²=AM²+MD²

2. AC²= AM²+MC²

3. AB²= AM²+MB²

4.BC²= AB²+AC²

equation(2) +equation (3) - 2 × equation (1)

AB²+AC²-2 AD²= AM²+MB²+AM²+MC²-2 AM²-2 MD²

                          =MB² +MC²-2 MD²

                         =(MB+MC)²-2 MB × MC -2 MD²

                          = BC² -2(BD²-MD²)-2 MD²→→As, MB=(BD-MD), and MC=(MD+CD), also AD is median, so BD=CD

AB²+AC²-2 AD²= BC² - 2 BD²

                             $=BC^2-\frac{BC^2}{2}\\\\=\frac{BC^2}{2}$    ---------(4)

Similarly, AB²+BC²-2 BE²  $=\frac{AC^2}{2}$  ---------(5)

CB²+AC²-2 CF²  $=\frac{AB^2}{2}$  ---------(6)

Adding (4),(5) and (6)

$2(AB^2+BC^2+AC^2)-2(AD^2+BE^2+CF^2)=\frac{(AB^2+BC^2+AC^2)}{2}\\\\ As, BC^2=AB^2+AC^2\\\\ 4 BC^2 -2(AD^2+BE^2+CF^2)=BC^2\\\\  3 BC^2 =2(AD^2+BE^2+CF^2)$

Hence proved.

Answer 2

look that attachments out ..

hope it helps ;)