Existence of the orthocenter. Prove that the altitudes of a triangle are concurrent. Hint: Use the trigonometric form of Ceva’s theorem. For each angle in the numerator, find a congruent angle in the denominator. Be careful with the special case of a right triangle.
Start with triangle ABC, let AE, BF, and CG be the altitudes. As you can see in the GeoGebra construction, AEC and BFC are similar by angle-angle similarity since angles E and F are both right angles, and the two triangles share angle C. Likewise, triangles AEB and CGB, and BFA and CGA are similar. Therefore, CE/CF = AE/BF, BG/BE = CG/AE, and AF/AG = BF/CG. Ceva's Theorem, says that (AG/BG)(BE/CE)(CF/AF) = 1 if and only if the cevians are concurrent. Then we can substitute the values to get (CG/CG)(AE/AE)(BF/BF) = 1. Therefore, by Ceva's Theorem, the altitudes of any triangle ABC are concurrent.