Below is an acute triangle ABC and its circumcircle. I have constructed the three altitudes AD, BE, and CF and extended each to intersect with the circumcircle at points P, Q, and R.

Notice, I have found that .

Proof:

In order to complete the proof, we will need to show that triangle ARB is congruent to triangle AHB, triangle AQC is congruent to triangle AHC, and triangle BPC is congruent to triangle BHC. We will want to do this because we know that triangle AHB + triangle AHC + triangle BHC= triangle ABC.

Triangle ARB congruent to triangle AHB.

Angle BAC and angle BRC subtend the same arc, arc BC. This implies angle BAC= angle BRC. Angle AEB of triangle AEB and angle RFB of triangle RFB are both right angles, therefore triangle RFB is similar to triangle AEB by AA similarity.

Angle AFC = angle ABC because they subtend the same arc, arc AC. Angle RFA of Triangle RFA and angle BDA of triangle BDA are right triangles. So, triangle RFA is similar to triangle BDA by AA similarity.

Triangle AEB and triangle AHB share angle ABE, which we have established is equal to an angle in triangle RBF. Triangle RFB and triangle ARB share angle RBF. Triangle AHB and triangle ARB share the side AB. Triangle RFA and triangle ARB share angle RAF. Triangle BDA and triangle AHB share angle BAD. Therefore, triangle ARB is congruent to triangle AHB.

Similar arguments can be made to show that triangle BPC is congruent to triangle BHC and triangle AQC is congruent to triangle AHC.

Since triangle ARB is congruent to triangle AHB, triangle AQC is congruent to triangle AHC, and triangle , then area triangle ARB + area triangle AQC +area triangle BPC= area triangle ABC.

So,

1 + areaARB + 1 + area AQC + 1 + area BPC =

1 + 1 + 1 +(areaARB + area AQC + area BPC) = 3 + 1 = 4