Congruences and angles: geometric solution and analytical solution

Question

Problem: Let $\triangle ABC$ be a right triangle in $A$. Let $O$ be the center of the square $BCDE$ constructed on the hypotenuse, on the side opposite the vertex $A$. Prove that $O$ is equidistant from the lines $AB$ and $AC$.

Solution:

We draw perpendiculars to $$ and $$ passing through $$, and call the points of $$ and $$ respectively the points of intersection. Proving that $$ is equidistant from $$ and $$ is equivalent to proving that $\cong $. Let $\gamma=A\hat C B$ and $\beta=A\hat BC=90^\circ-\gamma$. Consider the triangles $$ and $$ and prove that they are congruent.

1. $ \cong $ because they are half of the diagonals $$ and $$ of the square, which are congruent.

2. $\hat{} \cong \hat{}$ because right angles.

3. $\hat{C}K \cong \hat{B}K$. In fact $\hat{C}K=\gamma+45°$ and $ \hat{B}K=180°-(\beta+45°)=180°-(90°-\gamma+45°)=\gamma+45°$. The right triangles $$ and $$ are therefore congruent by the hypotenuse-angle criterion, and consequently accordingly $\cong $. We observe that the previous proof holds in the case where $\beta \geq \gamma$. In the event that $\beta < \gamma$ is valid $\hat{B}H=\beta+45°$ and $O\hat C K=180°-(\gamma+45°)=180° − (90° − + 45°) = + 45°.$

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Question: Is there possibly an analytical solution?

Answer 1

Here the required analitical proof.

Take $A=(-r,0),B=(r,0),A(x_0,y_0),O=(0,-r)$

Line $AB: y_0x-(x_0+r)y+ry_0=0$

Line $AC: (x_0+r)x+y_0y-(x_0+r)r=0$

$$d(O;AB)=\dfrac{|y_0\cdot0+r(x_0+r)+ry_0|}{\sqrt{y_0^2+(x_0+r)^2}}$$

$$d(O;AC)=\dfrac{|(x_0+r)\cdot0-ry_0-r(x_0+r)|}{\sqrt{y_0^2+(x_0+r)^2}}$$

We are done.

Answer 2

Here's a proof without words of the fact in the question:

(but I'll add a few words anyway; just notice that $O$ is also the centre of the square $AA'A''A'''$)