Part 2

If the parallel segment that divides the triangle into two equal areas is drawn for each base, a smaller triangle is formed. What is the ratio of the area of the small triangle to the original?

Let's label some points of the triangle:

Let's label the sides as follows:  a = BC, b = AC, and c = AB.

Since FI divides triangle ABC into two equal areas and is parallel to AC, we know (by construction), that a/FB = sqrt(2).  In other words, FB = a / sqrt(2).  Since a = FC + FB, we know that FC = a - FB = a - a / sqrt (2).

Similarly, BG = a - a / sqrt (2).

Since a = CF + FG + GB, we know that FG = a - FC - BG = a - (a - a / sqrt (2)) - (a - a / sqrt (2)) = 2 a / sqrt (2) - a.

Similarly, we also know the following side lengths:

GC = FB = a / sqrt (2)

AE = DC = b / sqrt (2)

BI = AH = c / sqrt (2)

EC = AD = b - b / sqrt (2)

AI = HB = c - c / sqrt (2)

DE = 2 b / sqrt (2) - b

HI = 2c / sqrt (2) - c

Since triangles GA'F and BIF are similar, (why?) we have the following ratio:

FG / FB = A'G / IB

For simplicity, let's let A'G = y.

Then substituting the values for FG, FB, and IB into the previous ratio and solving for y, we get the following:

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

If we let B'D = x, by following a similar procedure, we see that

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

Now look at the segment DG.  If we let A'B' = z, then we have the following:

DG = x + y - z.

By construction, DG = c / sqrt (2).  Thus, we now have

z = x + y - c / sqrt (2).

Substituting the values for x and y, we see

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

To find the ratio of the area of the two triangles, first we need to find the ratio of the side lengths of the two triangles.  Thus, we need to find z / c.

z / c  = (2c (sqrt (2) - 1) - c / sqrt (2))/c

z  / c  = 2 (sqrt (2) - 1) - 1 / sqrt (2)

z  / c = 1/2 (3 sqrt (2) - 4)

Thus, the ratio of the areas of the two triangles is 

(z / c)² = 1/4 (34 - 24 sqrt (2).

Click here for a GSP sketch to verify this ratio.

To go to Part 3 of the problem, click here.