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Austria competition problems

Geometry Difficulty 4.4 AIME Prove it Austria

Let $ABCD$ be a square. The equilateral triangle $BCS$ is constructed on the exterior of the side $BC$ . Let $N$ denote the midpoint of the line segment $AS$ and let $H$ be the midpoint of the side $CD$ .
Prove: $\angle NHC = 60^{\circ}$ .

Solution

Let $P$ be the midpoint of $BS$ , see Figure 1.

Since triangles $\triangle SNP$ and $\triangle SAB$ are similar with factor $2$ , the segment $NP$ is parallel to $AB$ and half the length of the segment $AB$ . Therefore $NPCH$ is a parallelogram.

As $NP$ and $BC$ are orthogonal and $PC$ and $BS$ are orthogonal, we have $\angle NPC = \angle CBP = 60^{\circ}$ .

Thus we obtain $\angle NHC = \angle NPC = 60^{\circ}$ .

Figure 1: Problem 2

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