The triangle pairs $\triangle ABI \cong \triangle AFC,\ \triangle BCE \cong \triangle BHA,\ \triangle CAG \cong \triangle CJB$ are congruent by SAS, for example, the 1st pair has the sides AB = AF and AI = AC equal, because ABEF and CAIJ are both squares and the angles $\angle BAI = \angle FAC = \angle A + 90^o$ are also equal. Hence, the other corresponding angles of these congruent triangle pairs are also equal. Moreover, the triangle $\triangle ABI$ is obtained by rotating the triangle $\triangle AFC$ around the vertex A by the angle $\angle FAB = 90^o$ and similarly for the other 2 triangle pairs. Hence, the line pairs $IB \perp FC,\ EC \perp AH,\ GA \perp BJ$ are perpendicular to each other. The angles $\angle P_1Q_1R_1 = \angle P_2Q_2R_2$ are equal, because $\angle P_1Q_1R_1 = \widehat{(P_1Q_1, R_1Q_1)} \equiv \widehat{(BJ, FC)} =$ $= \widehat{(GA, FC)} \equiv \widehat{(P_2Q_2, R_2Q_2)} = \angle P_2Q_2R_2$ and similarly, the angles $\angle Q_1R_1P_1 = \angle Q_2R_2P_2,\ \angle R_1P_1Q_1 = \angle R_2P_2Q_2$ are also equal. Thus we established that the triangles $\triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$ are similar. Let K, L, M be the diagonal intersections of the squares ABFE, BCGH, CAIJ and X, Y, Z the intersections of the perpendicular line pairs $IB \perp FC,\ EC \perp AH,\ GA \perp BJ$. The quadrilateral AXCI is cyclic, because the angles $\angle AIX \equiv \angle AIB = \angle ACF \equiv \angle ACX$ of the congruent triangles $\triangle ABI \cong \triangle AFC$ are equal. Since the angle $\angle CXI = 90^o$ is right, the circumcenter of this cyclic quadrilateral is the midpoint of the chord CI, the diagonal of the square CAIJ. Hence, the circumcircle of the quadrilateral AXCI is identical with the circumcircle $(M)$ of the square CAIJ and the square vertex J also lies on this circle. Since AJ is a diameter of the circle $(K)$ and the angle $\angle AZJ = 90^o$ is also right, the point Z also lies of this circle. The power of the triangle vertex B to the circle $(K)$ is equal to $BX \cdot BI = BZ \cdot BJ$, which implies $\frac{BI}{BJ} = \frac{BZ}{BX}$ The right angle triangles $\triangle BXQ_1 \sim \triangle BZQ_2$, because the angles $\angle BXQ_1 = \angle BZQ_2 = 90^o$ are both right and the angles $\angle BQ_1X \equiv \angle P_1Q_1R_1 = \angle P_2Q_2R_2 \equiv \angle ZQ_2B$ of similar triangles $\triangle P_1Q_1R_1 \sim \triangle P_2Q_2R_2$ are equal. Consequently, $\frac{BQ_2}{BQ_1} = \frac{BZ}{BX} = \frac{BI}{BJ}$ But this means that the triangles $\triangle BQ_2Q_1 \sim \triangle BIJ$ are centrally similar with the similarity center B and thus the lines $Q_1Q_2 \parallel JI \parallel CA$ are parallel. In a similar way, we could show that the lines $P_1P_2 \parallel HG \parallel BC$ and the lines $R_1R_2 \parallel FE \parallel AB$ are also parallel. Let S be the concurrency point of the lines AL, BM, CK, lying (together with the centroid and the 1st Fermat point of the triangle $\triangle ABC$) on Kiepert's hyperbola. Since the angles $\angle AMC = \angle AYC = 90^o$ are both right, the quadrilateral AYCM is cyclic. Since N is the midpoint of the are AC of its circumcircle opposite to the vertex Y, the line YC bisects the right angle $\angle AYC$. Let $(M_1), (M_2)$ be 2 centrally similar circles on centrally similar diameters $IJ \sim Q_1Q_2$ with the similarity center B. Since the right angle triangle $\triangle IMJ$ with the circumcircle $(M_1)$ is isosceles with IM = JM, the line BM cuts the corresponding arcs $JI, Q_1Q_2$ of the circles $(M_1), (M_2)$ at their midpoints M, S', the right angle triangle $\triangle Q_1S'Q_2$ is also isosceles with $Q_1S' = Q_2S'$ and the angles $\angle Q_2Q_1S = \angle Q_1Q_2S' = 45^o$. Since the angles $\angle Q_1XQ_2 = \angle Q_1ZQ_2 = 90^o$ formed by the perpendicular lines $Q_1X \equiv FC \perp IB \equiv Q_2X$ and $Q_1Z \equiv JB \perp AG \equiv Q_2Z$ are both right, the points $X, Z$ both lie on the circle $(M_2)$. Since the angle $\angle AZK \equiv \angle Q_2ZS = 45^o$, this angle spans the same arc $Q_2S'$ as the angle $\angle Q_2Q_1S' = 45^o$ and as a result, the points $S' \equiv S$ are identical. Thus we proved that the triangle $\triangle Q_1SQ_2$ is isosceles with the angle $\angle Q_1SQ_2 = 90^o$ and $Q_1S = Q_2S$. In a similar way, we could prove that the triangles $\triangle P_1SP_2,\ \triangle R_1SR_2$ are also isosceles with the angles $\angle P_1SP_2 = \angle R_1SR_2 = 90^o$ and $P_1S = P_2S,\ R_1S = R_2S$. As a result, the triangle $\triangle P_2Q_2R_2$ is obtained from the triangle $\triangle P_1Q_1R_1$ by a rotation around the center S by the angle $\angle P_1SP_2 = \angle Q_1SQ_2 = \angle R_1SR_2 = 90^o$, which means that these 2 triangles are congruent. Yetti

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thats crazy, can't we use analytic?

Dear Mathlinkers, an article concerning the Vecten's figure and its developpement can be seeing on my site http://perso.orange.fr/jl.ayme , la figure de Vecten vol. 5 p. 40 Sincerely Jean-Louis