Altitudes and Circumcircles by Emily Kennedy Let ΔABC be any acute triangle, and label it as follows (see below for a diagram): Draw the three altitudes of ΔABC, and let D, E, and F be the intersections of these altitudes with the side of the triangle to which they are perpendicular. Finally, let P, Q, and R be the intersections of the altitudes with the circumcircle (other than A, B, and C). Using this GSP file, make a conjecture about the following sum: Can you find any pairs of congruent segments? Use the Calculate > Distance command to measure lengths of segments. It looks like Let's prove it! First, let's modify the sum a little bit. So if we want to show that , we could equivalently show that    (*) Let's try the second one. Draw in hexagon CPBRAQ and label some angles as shown: We know that two angles on a single circle which subtend the same segment are congruent (see Assignment 9 for a "proof"). and both subtend segment , so they must be congruent. Thus, c = a + b.    (1) Since and are both right, and the sum of the interior angles of a triangle is 180°, we have: = 90° - a    and    = 90° - b Furthermore, since is a straight angle, we have d + (90° - a) + (90° - b) = 180° d + 180° - a - b = 180° d = a + b.    (2) By (1) and (2), we have d = a + b = c so Also, , since both are right angles. And finally, , clearly. So by AAS congruence, we have Thus, By similar reasoning, we have and Because of the congruences we just found, we can rewrite (*) as:    (**) Let's look at a few areas. For the whole triangle, ΔABC, we have AreaΔABC = AB·CF = AC·BE = BC·AD since CF, BE, and AD are each altitudes of the triangle. Now let's break the area up into three pieces: We now have: AreaΔABC = Areaorange + Areayellow + Areagreen = AreaΔAHB + AreaΔAHC + AreaΔBHC = AB·FH + AC·EH + BC·DH So we have This is precisely what we wanted to show (see (**))! Q.E.D.