Question

$BE$ and $CF$ are medians of a triangle $ABC$ right angles at $A$. Prove that $4(B{E}^2+C{F}^2)=5B{C}^2$.

Answer 1

In the triangle $\angle BAC=90˚$

$BE$ and $CF$ are medians, therefore

$BF=AF=c,AE=EC=b,BC=a$

Using the Pythagoras theorem, we can say that

$4(c^2+b^2)=a^2$.... (i)

In triangle $AFC$

$(CF{)}^2=(AF{)}^2+(AC{)}^2$

$\Rightarrow (CF{)}^2=c^2+4b^2$

In triangle $ABE$

$(BE{)}^2=(AE{)}^2+(AB{)}^2$

$\Rightarrow (BE{)}^2=b^2+4c^2$

Now,

$(BE{)}^2+(CF{)}^2=5(c^2+b^2)$

But from (i), $c^2+b^2=\frac{a^2}{4}$

Therefore,

$(BE{)}^2+(CF{)}^2=5(c^2+b^2)=\frac{5a^2}{4}=\frac{5(BC{)}^2}{4}$

$\Rightarrow 4(B{E}^2+C{F}^2)=5B{C}^2$