PROBLEM: Given any acute triangle ABC and its circumcircle. Prove that the three chords formed by the extensiions of the three altitudes of AZ,BY,CX and their corresponding altittudes AM, BN, and CP have the following relationship:

THE GENERAL PROBLEM
Labels: AM, BN, and CP are altitudes. AZ, BY, and CX are altitude extennsions that form chords with the circumcenter. H is the orthocenter.

APPROACH: Given a problem which didn't seem very obvious from the beginning I chose to look at very specific cases where I knew additional information that could aid me is seeing a pattern.

SPECIAL CASE #1 - THE RIGHT TRIANGLE
This special case brings with it much information. We find that AB is both the altitude and its chord altitude extension, similarily CA is both the atitude and its chord altidue extension. This relationshiip occurs because angle BAC = 90 degrees. We also know that AM = MZ because BC is perpendicular to AZ and BC goes through the center of the circle, thus BC is a perpendicular bisector of AZ..

SPECIAL CASE #2 - THE EQUILATERAL TRIANGLE
Diagram #1	Diagram #2
(Diagram #1) The equilateral presents us with more information than we need but a few of the key facts are that all corresponding segments of the triangle are equal and that each triangle within the equilateral is congruent. I will leave the proof of this to the reader. Instead of algebraically proving this I will use geometric relationships. (Diagram #2) By adding a few new lines CZ and BZ to the original diagram I am able to use the circle property that all inscribed angles on the same arc are equal . Thus as in the example above, angle CAZ = CBZ and angle BAZ = BCZ.
Diagram #3	Diagram #4
(Diagram #3) Using the inscribed angle property, a few basics facts about equilateral and congruent triangles, I am able to prove that HM = MZ . Using the same arguement I am able to reproduce the same triangle relationship in the upper portion of the equilateral. (Diagram #4) Thus AS = SH = HM = MZ. Using the four equivalent segments we see that the ratio of AZ to AM = 4AS to 3AS = 4 to 3. This ratio can be found on all three altitudes.
Diagram #5	Diagram #6
Thus which simplifies to because of the four equal segments found through congruent triangles.

INSIGHTS: Using the above specific cases I was able to find some patterns in the process that will help me in the general problem:

• that by adding additional chords from the vertex of the triangle to the altitude extensions (diagram #5 above, new chords CZ, CY, BX, and XA) new relationships begin to appear in the newly formed triangles. The biggest fact used is that the distance from the orthocenter(H) to the foot of the altitude is always equal to the distance from the foot of the altitude to its extension on the circle. Thus in diagram #5 above, HP = XP, HM = MZ, and HN = NY. This is true for all acute triangles.
• although I have not included my investiagtion on an isosceles triangle, it brought forth the need for looking at similar triangles to get new information. The equilateral dealt with congruency, but as we turn to the more general case of any acute triangle we must focus on similarity .

FINAL CASE - THE GENERAL PROBLEM
Using the above insights I have added the lines (above right) to the original problem. This allows me to use congruent triangles to prove that HN = NY, HM = MZ and HP = PX. The equivilent segments simplify into . Continuing with the algebraic simplification reduces to , thus .
Using the final equation from the above cell, we see that the equation simplifies into an interesting relationship:  for the three altitudes of the traingle. This can be proven through the use of similar triangles, but do to the length and complexity I will simply state that it can be done and that the below relationships are key in solving it.