I tried approaching the problem by drawing two altitudes (right angles at the base) and a third line not guaranteed to be at a right angle, but going through their common point of intersection. The result is a set of six triangles. So far I have been unable to manipulate equations involving these angles to yield a proof that the third line also meets its base in a right angle.
But I found this picture of a circumircle on wikipedia, and I found this when googling "prove orthocenter:" [Unfortunately, it appears that the books result doesn't take you back to the same page, which makes the link worthless...]
The circumcircle is the circle which contains each of the three vertices of the triangle. Its construction proceeds by finding the center of the circumcircle, which is the point equidistant from all three vertices. For each pair of vertices, on the line connecting them, which will become a chord of the circle, first find the midpoint and then construct a line perpendicular to the chord and passing through the midpoint by the standard method.
Each of these lines is parallel to the altitude but shifted, except in the case of an equilateral triangle, where it is the same, or an isosceles triangle, where one of them is the same. All the points on each line are equidistant from the two vertices used to construct the chord.
The three lines cross at a unique point. There might seem to be a similar problem here as with the orthocenter, except that we know there is a circle which includes all three of the vertices, and that it must have a center which is equidistant from the each of them. Since O (the center) is equidistant from A and B as well as from B and C, it is the same distance from each of them.
Above is my drawing of a generic triangle, its chords and perpendicular lines (called perpendicular bisectors), and the circumcircle and its center O. For this discussion I have labeled the vertices with the Roman letters A, B and C, rather than the sides.
Now, draw a second triangle inscribed in the first, with its vertices as the midpoints (x, y and z). It appears that the four smaller triangles are quite similar, and we will prove that they are congruent (identical), as well as being similar to the parent ΔABC. Furthermore, and most amazing, the circumcenter of ΔABC is the orthocenter of Δxyz
Start with the three outer triangles: ΔxBz, ΔyzC and ΔAxy. Obviously, each contains an angle of the original triangle, but also each of their sides is one half the length of the original side. Therefore, they are congruent with ΔABC and equal to each other (because e.g. Ax = xB).
Now for the inner triangle xyz. Consider the angle yxz. Notice that the other two angles formed where line AB crosses through x are angles of the set of congruent triangles. Therefore, angle yxz is also a member of that set (because 180 - Axy - Bxz = BCA). The inner triangle is congruent with the others, it is just rotated 180°.
Furthermore, because Δxyz shares a side with ΔxBz as well as its angles, we know they are not only congruent but equal. The shared side also proves that Δxyz is rotated 180°. And that means that xz is parallel to AC, xy is parallel to BC, and zy is parallel to AB.
Since xO was constructed as perpendicular to AB, it is also perpendicular to yz, therefore it is an altitude of the triangle xyz. Therefore the circumcenter O of the triangle ABC is orthocenter of the inscribed triangle xyz.
Summary:
We find the midpoints of the sides of a triangle ABC and inscribe a new triangle xyz. All five triangles are similar, and the four smaller ones are congruent (equal). Most amazing, the circumcenter of the large triangle is the orthocenter of the small central one. Very cool.
