Monday Math 153

A few weeks ago, I demonstrated a proof that all three medians of any triangle are concurrent; in other words, that the centroid exists.

Now, how about demonstrating that the circumcenter and orthocenter exist; that is to say, showing that the three perpendicular bisectors of the sides are concurrent, and that the three altitudes are concurrent, for any triangle.

The case for the circumcenter is simple to prove using the perpendicular bisector therorem: a point P is equidistant from points A and B if and only if P lies on the perpendicular bisector of the line segment AB.

So, for triangle ∆ABC, let point P be the intersection of the perpendicular bisectors of sides AB and BC. Since P is on the perpendicular bisector of AB, the perpendicular bisector therorem tells us that AP=BP; and similarly, since P is on the perpendicular bisector of BC, then BP=CP. Thus, AP=CP, and so the theorem tells us that P is on the perpendicular bisector of AC as well. We also have AP=BP=CP, and therefore that a circle with radius equal to that distance with center P passes through A, B and C.

The orthocenter case is a bit less simple. Construct through each vertex of triangle ∆ABC a line parallel to the opposite side, and let us label the intersections of these lines D, E, F as in the figure below:

OrthocenterFig1

Since the lines are constructed to be parallel, we see that the quadrilaterals ABDC, BCEA and CAFB are all three parallelograms. Since opposite sides of a parallelogram have equal length, we see from parallelogram BCEA that AE=BC, and from parallelogram CAFB that AF=BC. Thus, AE=AF, and A is the midpoint of EF. It is clear analogously that B is the midpoint of DF and C is the midpoint of DE as well.

Now, let \stackrel{\longleftrightarrow}{AH} be the altitude of ∆ABC from vertex A. By definition, \stackrel{\longleftrightarrow}{AH}\perp\stackrel{\longleftrightarrow}{BC}, and since \stackrel{\longleftrightarrow}{BC}\parallel\stackrel{\longleftrightarrow}{EF}, we see \stackrel{\longleftrightarrow}{AH}\perp\stackrel{\longleftrightarrow}{EF} as well. But A is the midpoint of EF, and so \stackrel{\longleftrightarrow}{AH} is the perpendicular bisector of EF. Similarly, the altitudes of ∆ABC from vertices B and C are the perpendicular bisectors of DF and DE, respectively.

But we proved above that the perpendicular bisectors of the sides of any triangle are concurrent. Thus, the altitudes of ∆ABC, being also the perpendicular bisectors of the sides of ∆DEF, must be concurrent.

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One Response to “Monday Math 153”

  1. Monday Math 154 | Twisted One 151's Weblog Says:

    […] Just another WordPress.com weblog « Monday Math 153 […]

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