**Problem**: (Arbelos - Volume 5, Chapter 4 Cover Problem) Let $ ABC $ be a triangle and let $ A^{\prime} $ be a point on the circumcircle of $ ABC $. Denote $ H $ the orthocenter of $ \triangle ABC $ and $ H^{\prime} $ the orthocenter of $ \triangle A^{\prime}BC $. Prove that $ AHH^{\prime}A^{\prime} $ is a parallelogram.

**Solution**: Let $ O $ be the circumcenter of the two triangles and set it to be the origin. Let $ A, A^{\prime}, B, C $ denote the vectors to each of the points, respectively. Note that $ |A| = |A^{\prime}| = |B| = |C| $.

We claim that $ H = A+B+C $ and $ H^{\prime} = A^{\prime}+B+C $ (as vectors).

To prove this we only need to show that $ H-A $ is perpendicular to $ B-C $, and by symmetry everything else follows.

Assume $ H = A+B+C $. Then $ H-A = B+C $. Consider the dot product $ (B+C) \cdot (B-C) = B \cdot B - C \cdot C = |B|^2-|C|^2 = 0 $. But the dot product of two vectors can be zero iff they are perpendicular, and the result follows.

Now back to the problem. Since $ AH $ and $ A^{\prime}H^{\prime} $ are both perpendicular to $ BC $, they are clearly parallel.

Consider writing $ AA^{\prime} $ as the vector $ A^{\prime}-A $ and $ HH^{\prime} $ as the vector $ H^{\prime}-H $, which by our claim above is equivalent to $ (A^{\prime}+B+C) - (A+B+C) = A^{\prime}-A $. Since $ AA^{\prime} $ and $ HH^{\prime} $ are the same vector, they are going in the same direction and thus parallel. Then $ AHH^{\prime}A^{\prime} $ is a parallelogram, as desired. QED.

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Comment: A very powerful use of vector geometry, as it simplifies the problem immensely, especially with the common formula for the orthocenter.

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Practice Problem: Let $ G $ be the centroid of $ \triangle ABC $ and $ G^{\prime} $ be the centroid of $ \triangle A^{\prime}BC $. Prove that $ GG^{\prime} $ is parallel to $ AA^{\prime} $.

[...] We know and from here, so we’ll use these results. Also, by the definition of a midpoint. Using these three equations, we can solve for . [...]

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