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:

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 be the altitude of ∆*ABC* from vertex *A*. By definition, , and since , we see as well. But *A* is the midpoint of *EF*, and so 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.

### Like this:

Like Loading...

*Related*

Tags: Altitude, Circumcenter, Math, Monday Math, Orthocenter, Parallelogram, Perpendicular Bisector, Perpendicular Bisector Theorem, Triangle Centers

This entry was posted on July 7, 2014 at 2:15 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

July 14, 2014 at 12:08 am |

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