Triangle inscribed in a square thales

Question

Thanks for helping me solving this problem

Let $ABCD$ be a square such that $AB = 4$. $A$ a point on the side $[AB]$ such that $AE = 1$

The segments $[DE]$ and $[AC]$ intersect at point $H$

-Calculate the distance $HE$

My attempt was to use thales on the triangles $HDC$ and $HAE$ and i found : $\frac{HE}{HD}=\frac{HA}{HC}=\frac{AE}{DC}$ that means : $\frac{x}{HD}=\frac{HA}{HC}=\frac{1}{4}$ then i tried to calculate $HD$ using the rectangle triangle $DAE$ but i always have two unknowns.

Answer 1

You know $AD$ and $AE$, so you can calculate $DE$ using Pythagoras. Now you have $$DE=HD+x$$ $$\frac{x}{HD}=\frac14$$ This is enough to deduce $x$: just express $HD$ in terms of $x$ using the second equation, and substitute that into the first equation.

Answer 2

If we set up coordinates such that $A$ is the origin, we can easily find out (by treating $DE$ and $AC$ as graphs of functions) that $H=(\frac45,\frac45)$. Hence $HE=\sqrt{(1-\frac45)^2+(0-\frac45)^2}=\frac{\sqrt{17}}5$.

Answer 3

Let $h$ be an altitude in triangle $AEH$. Then we have $${4-h\over h} = {4\over 1}$$

Dose that help?