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Ratios of Altitude Segments of a

by

Josh Traxler Figure 1: Constructed acute triangle.

Say we construct the altitudes on an arbitrary acute triangle ABC. We label the intersections of the altitudes D, E and F. We label the orthocenter H. (See Figure 1) We would like to prove the following two statements: HD HE HF + + = 1 AD BE CF and AH BH CH + + = 2 AD BE CF Following this, we will determine if this also holds true for obtuse and right triangles. HD First let’s take a look at . AD is an altitude of the triangle. HD is a AD portion of that altitude. Speciﬁcally, from the side to the orthocenter. We also observe that both AD and HD are the heights of a triangle if we let the be BC. AD is the height of triangle, ABC. HD is the height of the triangle BCH. Now if we look at the other two division in the problem, we get similar results. If we consider AC to be the base, the the height of ACH is HE and the height of ABC is BE. Similarly for HF and CF. So it is apparent that if we multiply the numerator and denominator by their corresponding bases divided by two, we will recieve ABC in the denominator, and a second triangle in the numerator. This operation is valid because we are only multiplying by 1. HD ∗ BC/2 HE ∗ AC/2 HF ∗ AB/2 + + AD ∗ BC/2 BE ∗ AC/2 CF ∗ AB/2 (BCH Area(ACH) Area(ABH) = + + Area(ABC) Area(ABC) Area(ABC) Area(BCH) + Area(ACH) + Area(ABH) = Area(ABC) Now we observe that triangles BCH ACH and ABH are a partition of the triangle ABC. Therefore, the sum of partition triangle’s should be the area of ABC. Therefore, we have the same value in the demoninator and numerator. HD HE HF So + + = 1 AD HE CF Now let’s consider the second equation we would like to work with:

1 Figure 2: Constructed obtuse triangle.