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

Also, , since both are right angles.

And finally, , clearly.

So by AAS congruence, we have


By similar reasoning, we have


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
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

So we have

This is precisely what we wanted to show (see (**))!


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