**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

T_{1}
= area of HAB

T_{2}
= area of HBC

T_{3}
= area of HAC

Then we have

T = T_{1} + T_{2} + T_{3}

ð
(T_{1}
+ T_{2} + T_{3}) / T = 1

ð
(T_{1
}/ T) + (T_{2 }/ T) + (T_{3 }/ T) = 1

But we also have that

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

T_{1}
= ½ (AB)(HF)

T_{2}
= ½ (BC)(HD)

T_{3}
= ½ (AC)(HE)

Hence T_{1
}/ T = HF / CF, T_{2 }/ T = HD / AD, and T_{3 }/ T = HE /
BE. So

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

Now note that:

AH = AD – HD

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.