Presented By

Dana TeCroney

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

Consider
acute ÆABC 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 ÆABC). This is because
each quadrilateral includes ÆABC and one of the orange triangles. The sum of the areas of the orange
triangles will be equal to the area of ÆABC 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 ÆBAD) will be the compliments of these congruent angles
and hence will be congruent. This
is because all the red lines are altitudes, hence ÆBAD and ÆRAF 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 quadrilateralsÉ

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(ÆABC)

I
now want to divide by the area of ÆABC.

The
area of ÆABC can be found in three ways,

Area(ÆABC)
= ½mACmBE

Area(ÆABC)
= ½mBCmAD

Area(ÆABC)
= ½mABmCF

The
moment of triumph,

And
of course,

.

QED.