Ratios of Altitude Segments of a Triangle
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 triangles 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 base 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 Area(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 areas 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.
AH BH CH + + AD BE CF The ﬁrst observation we make is that its similarity to the ﬁrst equation. The demoninators are the same and the numerators are still a segment of the altitude. The numerator in the second equation is the portion of the altitude from the vertex to the orthocenter, whereas the ﬁrst equation is the portion of the altitude from the orthocenter to the side. Therefore, together they are the entire altitude. For example, AD is partitioned into HD and AH. Cleary, HD AH AD AD HD AH + = = 1 And therefore, − = . Similarly for the AD AD AD AD AD AD other terms in the equations. AH We can use this, coupled with our previous solution to determine what + AD BH CH + is equal to. BE CF AD BE CF We know that + + = 3. AD BE CF HD HE HF Since we showed that + + = 1 , we can subtract this equation AD BE CF AD BE CF from + + = 3 to recieve: AD BE CF AD BE CF DH EH FH + + − + + = 3 − 1 AD BE CF AD BE CF By combining terms with similar fractions we recieve: AD − DH BE − EH CF − FH + + = 2 AD BE CF And due to the partitioning of the altitudes mentioned earlier this is equiv- elent to: AH BH CH + + = 2 AD BE CF Let’s say that the triangle is actually obtuse, as in Figure 2. Obtuse Triangles The orthocenter on an obtuse triangle exists outside the triangle. Without loss of generality, we will assume that BAC is the obtuse angle, as in Figure 2.
2 Figure 3: Constructed right angle.
HD Now we see that AD is a only a portion of HD. Therefore, > 1, invalidating AD the ﬁrst equation. We also see that since H lies out outside the circle on the ray, DA, BE is BH only a portion of BH and CF is only a portion of CH. Therefore, > 1 and BE CH BH CH > 1. Therefore + > 2, invalidating the second equation. So, CF BE CF due to the orthocenter being outside the triangle on an obtuse triangle, these equations do not ﬁt an obtuse triangle. However, we might ﬁnd that a modiﬁcation of our equation will work. In HD HE HF the case where 6 BAC is obtuse, the equation, − − = 1 , may be AD BE CF valid. HD HE HF − − AD BE CF HD ∗ BC/2 HE ∗ AC/2 HF ∗ AB/2 = − − AD ∗ BC/2 BE ∗ AC/2 CF ∗ AB/2 Area(BCH Area(ACH) Area(ABH) = − − Area(ABC) Area(ABC) Area(ABC) Area(BCH) − Area(ACH) − Area(ABH) = Area(ABC) We see that in the case of our obtuse triangle, BCH is partitioned into the triangles ACH, ABH and ABC. Therefore we see that the numerator of our equation must be the area of the partition that remains from BCH after ACH and ABH are taken away: the area of triangle ABC. So, the numerator equals the denominator and our equation equals one, satisfying our modiﬁed formula for obtuse triangles. Right Triangles We see from Figure three that a right the orthocenter of a right angle is at the location of the the vertex whose angle measure is ninety degrees. Without loss of generality, we assume that B is this vertex. This implies that HD=0, HE=BE, HF=0, AH=AB, BH=0, and CH=CF. HD HE HF 0 BE 0 + + = + + = 0 + 1 + 0 = 1 AD BE CF AD BE CF
AH BH CH AD 0 CF + + = + + = 1 + 0 + 1 = 2 AD BE CF AD BE CF
3 It is clear that our equations also work for right angle, who have the ortho- center on a vertice. Therefore, our equations ring true for acute triangles and right triangles, but not obtuse triangles.