MHB Triangular Relations: Proving (AD)^2+(DE)^2+(AE)^2= \frac{2}{3}(BC)^2

[solakis1]

Given a right ABC trigon with the right angle at A and two points D,E on BC such that: (BD)=(DE)=(EC)

Prove: $$(AD)^2+(DE)^2+(AE)^2= \frac{2}{3}(BC)^2$$

 

[Opalg]

Given a right ABC trigon with the right angle at A and two points D,E on BC such that: (BD)=(DE)=(EC)

Prove: $$(AD)^2+(DE)^2+(AE)^2= \frac{2}{3}(BC)^2$$

[sp][TIKZ][scale=0.75]
\coordinate [label=below: $A$] (A) at (12,0) ;\coordinate [label=below: $B$] (B) at (0,0) ;
\coordinate [label=above right: $C$] (C) at (12,5) ;
\coordinate [label=above left: $D$] (D) at (4,5/3) ;
\coordinate [label=above left: $E$] (E) at (8,10/3) ;
\coordinate [label=above: $N$] (N) at (1728/169,720/169) ;
\draw (D) -- (A) --node[below] {$c$} (B) --node[above left] {$a$} (D) --node[above left] {$a$} (E) -- (C) --node

{$b$} (A) -- (E) ;
\draw[dashed] (A) -- (N) ;
\draw (9.8,4.4) node {$a$} ;
\draw (1,0.2) node {$\theta$} ;[/TIKZ]
As in the diagram, let $AB = c$, $AC = b$, $BC = 3a$ and $^\angle ABC = \theta$. By the cosine rule, $$AD^2 = a^2 + c^2 - 2ac\cos\theta, \\ AE^2 = a^2 + b^2 - 2ab\sin\theta.$$ Then $AD^2+DE^2+AE^2 = 3a^2 + b^2 + c^2 - 2a(c\cos\theta + b\sin\theta).$ But by Pythagoras $b^2 + c^2 = (3a)^2 = 9a^2$. Also, if $AN$ is the perpendicular from $A$ to $BC$ then $$c\cos\theta + b\sin\theta = BN + NC = BC = 3a.$$ Therefore $$AD^2+DE^2+AE^2 = 12 a^2 - 2a(3a) = 6a^2 = \tfrac23(9a^2) = \tfrac23BC^2.$$[/sp]​