Assignment 8:

Altitudes and Orthocenters

by

Ángel M. Carreras Jusino

Goal:

Given a triangle

ABC, construct the orthocenterH. Let the pointsD,E, andFbe the feet of the perpendiculars fromA,B, andCrespectively. Show that:

ExplorationIn the following applet is shown

- the triangle
ABC- its orthocenter
H- the points
D,E, andFwhich are the intersection of the altitudes with the lines containing the sides of the triangle- a circle with diameter
BC- two lines perpendicular to
BCpassing throughBandCrespectively.Drag the point

Aand see what happens toH. Investigate what happen whenAis inside, outside, and on the circle; and inside, outside, and on the parallel lines. Also, check how the results of the two formulas vary as you moveA.

You should note that:

- When
Ais in the circle or outside the parallel lines, the triangle is obtuse andHlies outside of it.- When
Ais in the interior of the parallel lines and outside the circle, the triangle is acute andHlies inside of it.- When
Ais in the circle or in one of the parallel lines, the triangle is right andHlies on it, specifically in one of its vertices.- The results of the two formulas are constant (1 and 2 respectively) when the triangle is acute, i.e.,
His inside the triangle.

Case 1.ABCis an obtuse triangle.If

ABCis an obtuse triangle, then the orthocenter is outside the triangle.

Note that in the case of an obtuse triangle most of the following ratios are greater than 1.

In the diagram presented is clear that the ratios

are greater than 1.

Therefore for an obtuse triangle

Case 2.ABCis an acute triangle.Let consider the following figure to see if the given formulas are true for all acute triangles.

Remember from the

Explorationthat to have an acute triangle (and as a result haveHinside the triangle),Amust be in the interior of the two lines perpendicular toBCatBandC, and outside the circle with diameterBC. In terms of the figure above these can be summarized by the following expressions.(1)

(2)

We only consider the upper hemisphere of the circle in this case.

Note that the expression inside the radical is always positive because of (1).

*These facts [(1) and (2)], in addition to the two which are presented below, will be constantly used during the proof.*Now, we are going to find expressions for the distances required to prove the formulas algebraically.

Note that

First, let find the coordinates of the points

H,E, andFWe already know the equation of the altitude that goes through the vertex

A

So if we find the equation of another altitude, we can find the coordinates of the orthocenter (

H).Let work with the altitude from

Bto the line containing the segmentAC. The slope of the segmentACis given by

Now the slope of the altitude is given by

Because the altitude contains the origin, the equation of the altitude is

Now we find the interception of the altitudes

Therefore the orthocenter has coordinates

To find the coordinates of

E, we need to find the interception of the altitude fromBto the line containing the segmentACand the segmentAC. We already determined the equation for the altitude

Let look for the equation of the line containing

AC.The slope of this line is given by

This line contains the point

C, using the point-slope form we get

To find

Ewe solve the following system

ThereforeEhas coordinates

Similarly,Fhas coordinates

Now, let calculate the distances needed to prove the identities.

Now, let check the first identity

To check the second identity we need to find

AH,BH, andCH.

Now, let check the second identity.

Case 3.ABCis a right triangleLet consider the following figure.

Is clear that the altitudes from the perpendiculars from A and C coincide in B, therefore D, F, and H lie in the origin.

Now let find the coordinates of the point

E. This point lies in the intersection of the lineACand the altitude that goes through the vertexBFirst, the slope of the line

ACis given by , therefore its equation is .Because

BEis perpendicular toACits slope is , therefore its equation is .Solving the system we can find the coordinates of

E.

Ehas coordinates .Now, having all the coordinates of the points let proceed with finding the distances required to check the formulas.

.

Finally,

After the examination of these three cases we can conclude that given a triangleABC, with orthocenterH, and the pointsD,E, andFthe feet of the perpendiculars fromA,B, andCrespectively then if the triangleABCis not obtuse.