Solution to Earlier Problem with Two Equilateral Triangles
Monday 18 May 2026 at 22:15
Visible to anyone in the world
Edited by Richard Walker, Monday 18 May 2026 at 23:25
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.
Solution to Earlier Problem with Two Equilateral Triangles
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.