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This is a solution to the problem I posted 16 May 2026.

In the diagram triangles ABC and CDE are equilateral, with points A, C and E lying on a straight line. The problem is to prove CP and CQ have the same length.

There are probably many proofs - for example using coordinate geometry or complex number - but here is a short one using Euclidean geometry.

In the second diagram the coloured triangles ACD and BCE are congruent ('two sides and the included angle'), because AC = BC, CD = CE, and angle ACD = 120° = angle BCE . The two angles marked are therefore equal.

In the third diagram the coloured triangle CPD and the shaded triangle CQE are congruent ('two angles and the included side'), because angle PCD = 60° = angle QCE, angle PDC = = angle QEC and side CD = side CE.

Consequently CP = CQ which was to be shown.

Suppose we take a rectangle and erect equilateral triangles on two of its side, as shown in Fig. 1. Show that pointsC, E and B are the vertices of an equilateral triangle.

Here is my proof: If we draw in the sides of the triangle (Fig. 2) we can see triangles DCE, BEF and DFA all have

a side of length | and one of length ||
an angle of 150 degrees included between those sides.

Consequently the three triangles are all congruent (the same as on another) and DE = EF = FD, so DEF is equilateral as required.