# Bride's Chair

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The applet illustrates *Vecten's configuration* associated with a triangle ABC. The construction adds three squares formed on the sides of the triangle - either externally or internally. In case where the triangle is right-angled and the squares are external to the triangle, the configuration is fondly known as the *Bride's Chair*. The configuration has several engaging properties. First of all, the "add-on" triangles AB_{a}C_{a}, A_{b}BC_{b}, A_{c}B_{c}C have the same area as ΔABC. To see this, rotate (drag the slider), e.g., ΔAB_{a}C_{a} through 90° clockwise till B_{a} coincides with C. Then CA will serve as a median of ΔBCC'_{a} that splits the triangle into two of equal area. Also, after the rotation, the median AM_{a} will become a midline in ΔBCC'_{a}, parallel to its side BC, which implies the second property, viz., the same line serves as a median in ΔAB_{a}C_{a} and an altitude in ΔABC. It also has been observed that AM_{a} = BC/2 and similarly for the other medians.

The triangles AB_{a}C_{a}, A_{b}BC_{b}, A_{c}B_{c}C are known as *flanks* of ΔABC. The relationship is symmetric: a triangle is a flank of its own flanks. Thus, for example, we can also claim that the same line serves as an altitude in ΔAB_{a}C_{a} and a median in ΔABC. (Flank triangles have additional features.)

The configuration of a triangle and squares constructed externally on two of its sides appeared in a problem offered at the 54^{th} Leningrad Mathematical Olympiad (1988) for grade 9. It was required to prove that if the line A_{c}B_{c} is parallel to AB then ΔABC is isosceles:

The solution is easy. If A_{c}B_{c}||AB, then the altitude from vertex C in ΔABC extended is also the altitude in ΔA_{c}B_{c}C, meaning that in this triangle the same lines serves as the altitude and and the median from vertex C. This is true for both triangles A_{c}B_{c}C and ABC which implies that both are isosceles.

## Reference

- D. Fomin, A. Kirichenko,
*Leningrad Mathematical Olympiads 1987-1991*, MathPro Press, 1994

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