# Two Equilateral Triangles

The following problem was offered at the LXI Moscow Mathematical Olympiad for grade 9 (which is likely to be equivalent to the US grade 11):

In an equilateral triangle ABC, point D lies on BC and an equilateral triangle ADE is constructed on AD so that E is on the same side of AD as C. Prove that CE||AB.

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Solution

In an equilateral triangle ABC, point D lies on BC and an equilateral triangle ADE is constructed on AD so that E is on the same side of AD as C. Prove that CE||AB.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

A simple solution requires shifting the view point. Think of ΔADE as obtain not just by any construction but by rotating segment AD 60° counterclockwise. (This is assuming that ΔABC is oriented positively, i.e., AC is obtained from AB by a rotation through 60° in the counterclockwise direction.) It is noteworthy that in both cases the rotation is the same, viz., through 60° in the counterclockwise direction.

Once a rotation came into the picture, let's see what happens to other present elements under this rotation. So, B goes into, say, B' which is C, and C goes into C'. The whole segment BC maps onto the segment B'C',B'C',AD,AE,AB, so that C = B'. Point D is carried on onto point D' on CC'. So that AD' = AD,BC,AD,AB,DC. Furthermore, ∠DAD' = 60°. It follows that ΔADD' is equilateral,isosceles,scalene,obtuse,right,equilateral. In other words, D' = E. And we are done because CC' = B'C' is obtained from BC by a rotation through 60° while line AB is obtained from line BC by a rotation (around) A through 120° in the opposite direction.