$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. Find $AE : EC$ .

Question

$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. A circle with diameter $AD$ intersects the segment $AC$ at $E$. Find $AE : EC$.

What I Tried: Here is a picture :-

The first thing I did was use similarity. For example one can find multiple similar triangles like $\Delta ABC \sim \Delta ACD \sim \Delta CBD$. After assuming $AC = x$ and $AD = y$ we get everything there in the picture.

Next from Pythagorean Theorem we will get only one equation involving $x$ and $y$ :- $$5y^2 = x^2$$

The problem is, how will I proceed finding $AE$ and $ED$ ? I am thinking that $ED \parallel BC$ , but I have no proof of it and neither I cannot proceed further.

Can anyone help me? Thank You.

Answer 1

Note that $\angle AED$ must be $90^{\circ}$ as it is inscribed in a semi-circle. Thus $\triangle ADE \sim \triangle ABC$ and so

$$\frac{AE}{EC} = \frac{AD}{DB} = 4$$

Answer 2

$\angle DEA = 90º$ by Thales' theorem, so $\angle CED = 90º$.

Thus $\Delta DCE \sim \Delta CBD$ by AA. This gives:

$$\frac{EC}{CD} = \frac{DB}{DC} \Rightarrow \frac{EC}{2y} = \frac{y}{x}.$$

Write out $AC:EC$ for convenience in terms of $x,y$. Now using the fact that $5y^2 =x^2$:

$$\frac{AC}{EC} = \frac{2x}{2y^2/x} = \frac{x^2}{y^2} = 5.$$

Therefore $AE:EC$ is:

$$\frac{AC-EC}{EC} = \frac{5-1}{1} = \boxed{4}.$$