Altitudes and the Circumcircle: A Proof

by Kristy Hawkins


Before we can prove anything, we must know what it is that we are proving. Begin by constructing any acute triangle ABC and its circumcircle. Then construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.

When measuring the lengths of certian segments in GSP, we can find,

Why did this happen? Let's try to prove that this is true for all acute triangles!

To begin with, we can simplify the equation by breaking up AP, BQ, and CR. We can see that AP=AD+DP, BQ=BE+EQ, and CR=CF+FR. Therefore,

This make our proof more simple! Now this last statement is all that we have to prove. To continue the proof, let's look at the area of triangle ABC. This can be expressed in 3 different ways since we have 3 different altitudes.

If we substitute these equations into ours we will simplify it to

Now if we can show that the numerator of this fraction is equal to 2a, then we have proven our original claim.

To do this, we will look to congruent triangles. If we can prove that triangle AHF is congruent to triangle ARF, then we will be in good shape. We see that both triangles share segment AF. Also angle AFH is congruent to angle AFR since they are both right angles by construction. If we show that angle FAH is congruent to angle FAR, then we have shown that our two triangle are congruent.

Angle FHA is congruent to angle DHC since they are vertical angles. This means that angle BAP must be congruent to angle BCR because of the right triangle congruence theorem. We see that angle RCB corresponds to chord RB. Then we have a picture like this.

Any two angles corresponding to the same chord on a circle are congruent, therefore angle RAB is congruent to angle RCB. This proves that triangles AFR and AFH are congruent by the ASA congruence theorem.

Since we have shown these two triangles congruent, we know that in the same way triangle BHD is congruent to triangle BPD and triangle CHE is congruents to triangle CQE. We can now express the area of triangle ABC in terms of triangles AHB, AHC, and BHC.

Therefore we have shown that

and hence