Bride's Chair
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 ABaCa, AbBCb, AcBcC have the same area as ΔABC. To see this, rotate (drag the slider), e.g., ΔABaCa through 90° clockwise till Ba 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 AMa 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 ΔABCa and an altitude in ΔBCC'a. It also has been observed that AMa = BC/2 and similarly for the other medians.
The triangles ABaCa, AbBCb, AcBcC 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 ΔABaCa 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 54th Leningrad Mathematical Olympiad (1988) for grade 9. It was required to prove that if the line AcBc is parallel to AB then ΔABC is isosceles: AC = BC.
The solution is easy. If AcBc||AB, then the altitude from vertex C in ΔABC extended is also the altitude in ΔAcBcC, 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 AcBcC 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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