The Ratio of the Segments Created by the Orthocenter

by

Susan Sexton

Given triangle ABC, construct the orthocenter H. Let points D, E, and F be the feet of the altitudes from A, B, and C respectfully.

Here I will show that

and

I. Proof of:

Let us first note the following:

HD is the height of ∆BHC with base BC.

HE is the height of ∆CHA with base AC.

HF is the height of ∆AHB with base AB.

AD is the height of ∆ABC with base BC.

BE is the height of ∆ABC with base AC.

CF is the height of ∆ABC with base AB.

So we can alter the equation by multiplying each element by a form of 1 as done below:

Now note that

Area of ∆BHC is:

Area of ∆CHA is:

Area of ∆AHB is:

Area of ∆ABC is:

and

Area of ∆ABC is also:

So we can substitute into our newly formed equation:

II. Proof of:

Let us first note the following:

AH = AD – HD

BH = BE – HE

CH = CF – HF

So by substitution we have:

(from the proof above)

What happens if ∆ABC is an obtuse triangle?

As angle ABC gets closer to 180 then any segment whose endpoint is H (HE, HB, HA, HD, HC, HF) will grow larger and segments AD, BE, and CF will get smaller.

So we will have:

therefore

So the first proved statement does not hold.

Additionally:

therefore

and

So the second proved statement does not hold.

Here is a GSP sketch to see how the numbers change when ∆ABC changes.

Orthocenter Ratio

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