The Art and Craft of Problem Solving

268 CHAPTER 8 GEOMETRY FOR AMERICANS

and BD intersect outside the triangle (remember, they are lines, not line segments).

A

�c

/

/ /

�E

How do we show that the three altitudes meet in a single point? Well, what other lines

are concurrent? The perpendicular bisectors are almost what we want, and they meet

in a single point, the circumcenter. However, altitudes are perpendicular to opposite

sides. In general, they do not bisect them. But is there any way to make them bisect

something? Yes! The ingenious trick is to make the point of bisection be the vertex,

not the side.

In the figure below, we start with an arbitrary triangle ABC, and then draw lines

through each vertex that are parallel to the opposite sides. These three parallel lines

form a larger triangle, E F D, that contains ABC.

F

Let m be the altitude of triangle ABC that passes through B. Certainly, m is per­

pendicular to AC, by definition. But since DE II AC, then m is also perpendicular to

DE.

Notice that ADBC is a parallelogram, so AC = BD by Example 8.2.9. Likewise,

AC = BE. Thus B is the midpoint of DE, so m is the perpendicular bisector of DE.

By the same reasoning, the other two altitudes of triangle ABC are perpendicular

bisectors of EF and F D. Consequently, these three altitudes meet in a point, namely

the circumcenter of triangle DEF! •