Part 3

What is the ratio of the area of the shaded triangle to the area of the original triangle in the figure below? Here again the segments parallel to the bases divide the original triangles into two equal areas.

First let's label some points:

From part 2, we know that HI = 2 c / sqrt (2) - c, where c = AB.  Thus,

HI = c (sqrt (2) - 1).

Now we need to find the ratio of the side lengths of the two triangles.  

HI / c = c (sqrt (2) - 1) / c.

HI / c = sqrt (2) - 1.

The ratio of the areas (since the two triangles are similar), then, is

(HI / c)² = (sqrt (2) - 1)² = 3 - 2 sqrt 2.

Click here for a GSP sketch to verify this ratio.

Similarly, the ratio of the area of triangle DB'E to triangle ABC is 3 - 2 sqrt 2, as is the ratio of triangles A'GF and ABC.  Thus, triangles AHC', DB'E, and A'GF have equal areas.

To go to Part 4 of the problem, click here.