Altitudes and Orthocenter

By Thuy Nguyen

Theorem:

Given triangle ABC, let H be the orthocenter and let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Then

Proof:

Let

T = area of ABC

T1 = area of HAB

T2 = area of HBC

T3 = area of HAC

Then we have

T = T1 + T2 + T3

ð    (T1 + T2 + T3) / T = 1

ð    (T1 / T) + (T2 / T) + (T3 / T) = 1

But we also have that

T = ½ (AB)(CF) = ½ (BC)(AD) = ½ (AC)(BE)

T1 = ½ (AB)(HF)

T2 = ½ (BC)(HD)

T3 = ½ (AC)(HE)

Hence T1 / T = HF / CF, T2 / T = HD / AD, and T3 / T = HE / BE. So

(HF / CF) + (HD / AD) + (HE / BE) = 1.

Now note that:

BH = BE – HE

CH = CF – HF

Substituting this into (AH / AD) + (BH / BE) + (CH / CF) we get

(AD – HD) / AD + (BE – HE) / BE + (CF – HF) / CF = 3 – [(HD / AD) + (HE / BE) + (HF / CF)] = 3 – 1 from previous result

= 2. QED.

Now in the case when triangle ABC is obtuse, the results no longer hold since the orthocenter H lies outside of our triangle. But if we consider the triangle HBC, then our orthocenter is at A and the results shown above will hold.

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