Question

Q.) AD , BE and CF are medians of the ABC , then prove that

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

Answer 1

Answer on Question #46827 – Math – Geometry

Q.) AD, BE and CF are medians of the ABC, then prove that

$$3(\mathrm{AB2} + \mathrm{BC2} + \mathrm{CA2}) = 4(\mathrm{AD2} + \mathrm{BE2} + \mathrm{CF2})$$

Solution.

$$AB^2 = h^2 + (x - a)^2, \quad AC^2 = h^2 + (x + a)^2 \rightarrow$$ $$AB^2 + AC^2 = h^2 + (x - a)^2 + h^2 + (x + a)^2 = 2(h^2 + a^2 + x^2) =$$ $$= 2(h^2 + a^2) + \frac{(2x)^2}{2} = 2AD^2 + \frac{1}{2}BC^2. \text{ So,}$$ $$AB^2 + AC^2 = 2AD^2 + \frac{1}{2}BC^2 \quad (1)$$

Similarly,

$$BC^2 + AC^2 = 2CF^2 + \frac{1}{2}AB^2$$ $$AB^2 + BC^2 = 2BE^2 + \frac{1}{2}AC^2$$

Adding (1), (2), and (3) we have:

$$2(AB^2 + BC^2 + AC^2) = 2(AD^2 + CF^2 + BE^2) + \frac{1}{2}(AB^2 + BC^2 + AC^2)$$

Or,

$$3(AB^2 + BC^2 + AC^2) = 4(AD^2 + CF^2 + BE^2) \quad \text{Q.E.D.}$$

www.AssignmentExpert.com