$ABCD$ : square and $E\in BD$ prove that $CE⊥PQ$

Question

Problem :

Let the square $ABCD$ and $E$ point in $BD$

and $P_{AD}(E)=J$ , $P_{AB}(E)=H$ ($P$ : projection orthogonal of E ...)

Now we need prove that $CE\perp JH$ ( perpendicular ) for any point $E$ in $BD$ My idea take a $(D,DC,DA)$

Then I will search cordone of point $E,C,H,J$

$D=(0,0),A=(0,a),C(a,0)$ but I don't how I find cordone of $E$ I need see other method if exist Can we use barycenter here ?

Answer 1

The red segments are diagonals of congruent rectangles, which can be overlapped through a $90^\circ$ rotation around their common vertex, followed by a translation. It follows that the red segments are orthogonal.

Answer 2

As stated in the comments, I would rather use a purely geometrical approach:

Hint: Show that the green and blue triangles are similar.

It's easy to take it from here since $JE\parallel DC\implies \angle IEJ=\angle ECD\implies...$

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For the analytical approach: Take $E(e,e)$ considering that $0\leqslant e\leqslant 1$. Therefore $H(e, 1)$ and $J(0, e)$. Now, it is just a matter of evaluating slopes. Observe that $$m_{HJ}=\frac{1-e}{e}\qquad m_{CE}=-\frac{e}{1-e}\qquad \implies \qquad m_{HJ}\cdot m_{EC}=-1$$

Answer 3

I like analytical approach better: $J$ is at $(0,x)$, $H$ is at $(x,a)$. The slope of $CE$ is $$m_{CE}=\frac{0-x}{a-x}$$The slope of $JH$ is $$m_{JH}=\frac{a-x}{x-0}$$ Multiply the two and you get $$m_{CE}\cdot m_{JH}=-1$$ So they are perpendicular