Proving that a line is an angle bisector

Question

In a right triangle $\triangle ABC$ with $A=90^{\circ}$ we inscribe a square which one side of it is located on hypotenuse $BC$. Prove that the line which joins vertex $A$ to the center of square is angle bisector of angle $A$.

My attempt: if we call the center of square $O$ we have to prove angle $BAO$ and the angle $CAO$ are $45^{\circ}$.

Answer 1

Let $O$ be the center of the inscribed square. Moreover let $D$ be the vertex of the square along $AB$ and let $E$ be the vertex of the square along $AC$. Then $DE$ is parallel to $BC$ and, by the law of sines applied to the triangles $\triangle ADO$ and $\triangle AEO$, we have that $$\frac{|DO|}{\sin(DAO)}=\frac{|AO|}{\sin(B+45^{\circ})} =\frac{|AO|}{\sin{(C+45^{\circ}})}=\frac{|EO|}{\sin(EAO)}$$ where the second equality holds because $B+C=90^{\circ}$. Note that $|DO|=|EO|$, hence it follows that $\sin(DAO)=\sin(EAO)=\cos(DAO)$, that is $DAO=EAO=45^{\circ}$.

Answer 2

Let $D$ be the center of the square $PQRS$, where $P\in AB$ and $Q\in AC$.

Thus, $DPAQ$ is cyclic and since $PD=QD$, we obtain $$\measuredangle DAP=\measuredangle DAQ$$ and we are done!