The Relationship Between Altitudes and the Circumcircle

by: Lauren Wright

with the assistance of Andy Tyminski

In this investigation, we will begin with an acute triangle ABC and its circumcircle.

Next, we will construct the three altitudes AD, BE, and CF.

Lastly, we will extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.

Now, let's find .

From our GSP calculations, = 4.

Is this true for any acute triangle? Click here to manipulate this sketch.

Proof

1. We know that there are three pairs of similar triangles here, the yellow, the teal, and the orange.

2. After drawing in some auxiliary lines, we know that triangle DHC is congruent to triangle DPC; triangle AHE is congruent to triangle AQE; and triangle AHF is congruent to triangle ARF. To see a proof of these congruencies, click here.

So, by definition of congruency: HD = PD, HF = RF, and HE = QE.

Now, we move on to the actual proof...

1. Let's look at each ratio individually:

2. By substitution, we get the following:

3. To show that , we will have to show that , or that:

.

4. In order to do this, we are going to examine the area of triangle ABC and its relationship to the areas of triangles BHC, AHC, and AHB.

We are going to look at twice the area of triangle ABC in order to minimize our arithmetic.

Let's try to solve for since it is the first ratio in .

5. Since the right hand side of each equation is equal to , we will take the two values that contain DH and AD and set them equal to each other. Then we will solve for .

1.

2.

3.

4.

Note: and .

5.

6.

7.

8.

6. CONCLUSION

Since , we can conclude that .