Presented By

Dana TeCroney

 

The purpose of this assignment is to explore some relationships between circumcircles and altitudes.

 

Consider acute ∆ABC with its circumcircle drawn, and altitudes AP, BQ, and CR.

What I wish to prove is that .

 

To prove this, first draw in cords BP and CP.

 

The area of quadrilateral ABPC = ½mBDmAP + ½mDCmAP = ½mBCmAP.

 

Also, the area of quadrilateral ABCQ = ½mACmBQ,

and the area of quadrilateral ARBC = ½mABmCR  

 

If it can be shown that the yellow and orange triangles below are congruent, then the sum of the areas of the three quadrilaterals ABPC, ABCQ, ARBC will be equal to 4(area ∆ABC).  This is because each quadrilateral includes ∆ABC and one of the orange triangles.  The sum of the areas of the orange triangles will be equal to the area of ∆ABC if the orange and yellow triangles are congruent.

First, I will prove the following two triangles are congruent.

 

Angles ARF and ABC are congruent since they subtend the same arc (AQC). Also, angles RAF and BAD (look at ∆BAD) will be the compliments of these congruent angles and hence will be congruent.  This is because all the red lines are altitudes, hence ∆BAD and ∆RAF are right triangles.  Since they share side AF, the implication of is that the yellow and orange triangles above are congruent by a side-angle-angle argument.

 

This can be extended to show that the highlighted triangles below are congruent since they share a common side AB, have congruent sides RA and AH, and congruent angles RAB and BAH.

 

 

A similar argument can be made to show the following sets of triangles are congruent.

 

         

 

Back to the quadrilaterals…

 

What can we say about the areas of the following quadrilaterals? 

 

Area(ABCQ) + Area(ABPC) + Area(ARBC) = 4*Area(ABC)

 

But, the area of quadrilateral ABPC = ½mAPmBC

the area of quadrilateral ABCQ = ½mACmBQ,

and the area of quadrilateral ARBC = ½mABmCR  

 

This implies ½mACmBQ + ½mBCmAP + ½mACmBQ = 4*Area(∆ABC)

 

I now want to divide by the area of ∆ABC.

 

The area of ∆ABC can be found in three ways,

Area(∆ABC) = ½mACmBE

Area(∆ABC) = ½mBCmAD

Area(∆ABC) = ½mABmCF

 

The moment of triumph,

 

 

And of course,

 

.

 

QED.