What is ratio of $BE : CE$?

Question

known : $\angle A = 90^{\circ}$. $AB=AC=2AD$ also $AE$ vertical to $BD$.

Let's say the side is $a$. And point $ O $ is the intersection point of $BD$ and $AE$. I found $AO$ $\frac{a}{\sqrt 5}$. I assume $\angle BOE$ and $\angle DOE$ is $90^{\circ}$. I draw the plane as . But how to find the ratio of $BE : CE$ ?

Answer 1

A coordinate geometry proof (Origin at $A$)

$BD$ has gradient $-2$ and so $AE$ has gradient $\frac{1}{2}$. The line $BC$ has equation $x+y=a$ and so $AE$ and $BC$ cross at $(\frac{2a}{3},\frac{a}{3})$.

The ratio is $2:1$.

Answer 2

Let $C$ be a mid-point of $AI$ and $AE\cap BI=\{G\}$.

Thus, since $\Delta ABD\sim\Delta AIB,$ we obtain: $$\measuredangle GAD=\measuredangle ABD=\measuredangle BIA,$$ which gives $$AG=GI=BG$$ and since $BC$ and $AG$ are medians of $\Delta ABI,$ we obtain: $$BE:EC=2:1.$$

Answer 3

$\begin{split} BE:CE & = \triangle ABE:\triangle ACE &(1)\\ & = \mathrm{d}(B,AE):\mathrm{d}(C,AE) &(2)\\ & = BO:(2DO) &(3)\\ & = \frac45\sqrt5AD:(2\times\frac15\sqrt5AD)\quad&(4)\\ & = 2:1 &(5) \end{split}$

1. The two triangles have bases $BE,CE$ and equal height from $A$.
   
2. The two triangles have the same base $AE$ and different height, where $\mathrm{d}(B,AE)$ denotes the distance from $B$ to $AE$, and so on.
   
3. $\mathrm{d}(B,AE)=BO$, clearly.
   $\mathrm{d}(C,AE):\mathrm{d}(D,AE)=AC:AD=1:2\Rightarrow \mathrm{d}(C,AE)=2DO$
4. and 5. Simple calculation. If you have any problem, feel free to comment below.