(i) Let A'' be the intersection point of the circumcircle (c) of ABC with the medial line of BC lying on the same side as A. Define {B'', C''} analogously.

(ii) On the arc A''A take a point A' such that the ratio of arcs A''A/A''A' = 3. Define {B', C'} analogously.

[1] The resulting triangle A'B'C' is equilateral.

[2] The oriented arcs satisfy AA' + BB' + CC' = 0.

Arc AA'' corresponds to a central angle of 2B-(pi-A) = B-C and consequently the central angle of AA' is (2/3)(B-C). Thus, the arc A'ABB' corresponds to a central arc of 2C + 2/3(B-C) + 2/3(A-C) = (2/3)(A+B+C) = (2/3)pi. This proves [1].

The second statement follows trivially from [1]. It is interesting to note that angle(A'OA'') is equal to the angle of lines (BC, B'C') since their sides are correspondingly orthogonal. Last angle corresponds to the algebraic sum of arcs BB'+CC'. To see this draw a parallel to BC from C' and measure the arcs along BB'.

This exercise is used in the discussion in Deltoid.html and a more general form of it is used in S_Triangles.html .

S_Triangles.html

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