The Art and Craft of Problem Solving

8.4 THE POWER OF ELEMENTARY GEOMETRY 291

Our strategy for proving that H, M, and K are collinear is to show that H M and
M K have equal "slopes." This demands that we find equal angles somewhere. We will
do so by finding similar triangles.
We drew in the medial triangle, because it "connects" points K and H in an intrigu­
ing way: the medial triangle is similar to ABC, and (have you done Problem 8.2.2 9
yet?) the circum center is the orthocenter of the medial triangle. In other words, K is

the orthocenter of D,.EGI, and D,.EGI "-' D,.ABC, with a ratio of similitude of I : 2.

Furthermore, the centroid of D,.EGI is also M, so at this point, we are nearly done.
Segments H M and KM are "sisters," each connecting the orthocenter and centroid of
similar triangles. Let's formalize this.

A

C

Draw a line through K perpendicular to IG. Since K is the orthocenter of D,.EGI and
the circumcenter of D,.ABC, this perpendicular line passes through E, the midpoint of
BC. Draw median AE, and we see that

LHAM = LKEM,

since AD II EK.
Finally, draw segments HM and MK. We'd like to show that they form a straight
line, which would be implied by D,.HAM "-' D,.KEM (which would then imply, for
example, LAHM = LEKM, so lines HM and MK have the same slope). We are al­
most able to prove this, since we have the angle equality; all we need are proportional
lengths. And this is easy. We have EK/AH = 1 /2, because D,.EGI "-' MBC. And
EM / MA = 1/2, by the centroid theorem (8.1 .3). So by "proportional SAS" (8.3.8) we
conclude that D,.HAM "-' D,.KEM, and we're done. _