Sinje J. Butler

The following is a sketch of triangle **ABC** with its orthocenter **H.**

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Take a look at triangles **HBC,
HAB, **and **HAC**. What will the location of their
orthocenters be?

Recall that the definition of the orthocenter of a triangle
is the common intersection of the three lines containing the altitudes and use
this definition to think through the process of determining the orthocenter for
triangle **HAC**.

The altitude from vertex **C**
to line segment **AH **is **a1
**on the graph below. We
know this because the line containing segment **AH**
contains one of the altitudes of
the original triangle **ABC, **thus, this
line is perpendicular to line segment **BC****,** further, the segment from the vertex **C** to the foot of the perpendicular on **BC** is perpendicular to the line
containing segment **AH****. **The
line containing this altitude runs through line segment** BC.**

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The altitude from vertex **H** to segment **AC
**is** a2** because the line
containing **a2 **contains one of the
altitudes of the original triangle.

It is now clear that the lines containing **a1 **and **a2**
intersect at vertex **B** of the
original triangle. Since the
orthocenter of a triangle can be determined by the common intersection of two
lines containing the altitudes, the orthocenter of triangle **HAC **is vertex **B** of the original triangle.

By using this same line of reasoning you can state that the
orthocenter for triangle **HBC**** **is vertex** A
**and the orthocenter for **HAB**
is vertex **C.**

Please click here for a **GSP demonstration**.
Notice for triangle **HAC**
with orthocenter at vertex **B, **that
when the orthocenter is outside the triangle, **HAC**
appears to be an obtuse triangle. **B **appears to lie on top of vertex **H** when the triangle is a right
triangle. And, finally when **B **moves into the triangle, the triangle
appears to be acute. This is how
we would expect the orthocenter to behave.

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