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,
Is this true for any acute triangle?
Click here to manipulate this sketch.
We start with what we already know.
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,
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 .
we can conclude that .