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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 $45^{\circ}$.

Michael Rozenberg
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max
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    Your “attempt” is just a restatement of the problem. – amd Jan 10 '18 at 08:16
  • @dxiv In the question, a side of the square is on the hypotenuse. Presumably your "vertex of the square on the hypotenuse" means where the extension of $AO$ meets $BC$ and you are looking at another square – Henry Jan 10 '18 at 08:42
  • @Henry Right, thanks for pointing out. Comment removed. – dxiv Jan 10 '18 at 16:40

2 Answers2

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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 $DAO=EAO=45^{\circ}$.

Robert Z
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  • How come you have equated law of sines of two different triangles? – max Jan 10 '18 at 11:37
  • @max Law of sines explains the first and the third equality. The second equality follows from the fact that $\sin{(B+45^{\circ}})=\sin{(C+45^{\circ}})$ since $B+C=90^{\circ}$. – Robert Z Jan 10 '18 at 11:44
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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!

Michael Rozenberg
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  • @ Rozenberg what do you mean by cyclic? Would you refer me to an article or something Please! – max Jan 10 '18 at 11:31
  • It means that there is a circle, which passes trough points $D$, $P$, $A$ and $Q$. Because $\measuredangle A+\measuredangle PDQ=180^{\circ}.$ – Michael Rozenberg Jan 10 '18 at 14:52