In the right triangle ABC where ∠B = 90◦
, BC : AB = 1 : 2. Construct the median BD
and let point E be on BD such that CE ⊥ BD. Determine BE : ED
(Figure below)
I already solved this problem, but the methods I used were really long and maybe there is a property that makes things a lot easier in this problem.
I already tried using coordinate geometry and trigonometry, the coordinate geometry is pretty straight forward so I'll only show the trigonometry solution
Trig Solution:
By Pythagorean Theorem, $DC = \frac{\sqrt5}{2}$, By Cosine Law on triangle CDB, We see that $BD = \frac{\sqrt5}{2}$.
We know that
$\sin (DCE) = \frac{2DE}{\sqrt5}$
$\sin (BCE) = BE$
$\sin(BCE + DCE) = \frac{2}{\sqrt5}$
Using these equations, we can find that
$CE(BE + ED) = 1$
then $CE = \frac{2}{\sqrt5}$
Using Pythagorean on CDE and CBE, we find that the ratio of the 2 lengths is 2:3.
Any solution that is different from these is appreciated.
Best Answer
Let $A(0,2)$, $B(0,0)$, $C(1,0)$, hence $D(1/2,1)$. The straight line through $BD$, $y=2x$, meets the line through $CE$, $y=-x/2+1/2$, for $x=1/5$, the first coordinate of $E$. Since the first coordinate of $D$ is $1/2$, it follows that $$BE:ED=\frac{1}{5}:\left(\frac12-\frac15\right)=2:3.$$