*Altitudes and the Orthocenter*

*By Sharon K. OÕKelley*

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__The Problem__

**Given
triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of
the perpendiculars from A, B, and C respectfully.**

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Prove É. **

**
ProveÉ.**

__Proof 1__

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**
Prove É. **

**Since
there are three different altitudes or height with three different
corresponding bases, the area of triangle ABC can be expressed in three waysÉ.**

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**Note that
within triangle ABC there are three smaller triangles – i.e., , , and The areas of
these triangles areÉ.**

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**The
areas of the smaller triangles and of triangle ABC can now be expressed as
ratiosÉ.**

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**Consider
that the three smaller triangles comprise triangle ABC; therefore, the sum of
their areas will equal the area of triangle ABC. This fact allows for the
following equationÉ.**

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**This
equation can be manipulated to produceÉ.**

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**Using
substitution, the desired result is obtainedÉ.**

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Q.E.D.**

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__Proof 2__

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**
ProveÉ.**

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**Consider
the result from Proof 1 thatÉ.**

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**In
triangle ABC, segments AH, BH, and CH are parts of their respective altitudes;
therefore, they can be used to express the above equation using a variation of
the segment addition postulateÉ.**

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**This
equation can now be rewritten asÉ.**

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**This
can now be manipulated to obtain the desired resultÉ.**

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** ****Q.E.D.**

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__What if the Given Triangle
is Obtuse? Will These Two Results Hold?__

**It
was demonstrated in Proof 1 that when triangle ABC is acute thatÉ.**

**It
follows then thatÉ.**

**This
also makes sense because segment HE is a part of segment BE; therefore, it is
less than the whole.**

**Now
consider when triangle ABC is obtuse and the orthocenter is outside the
triangle as shown in the diagram below.**

**Now
segment BE is a part of segment HE thereforeÉ.**

**This
would mean thatÉ.**

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**Thus,
the results from Proof 1 and thereby Proof 2 do not hold for the obtuse case****.**

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