3-2 In \triangle ABC, median \overline{AD} is perpendicular to median \overline{BE} (Fig. S3-2).
Find AB if BC = 6 and AC = 8.
Let AD = 3x; then AG = 2x and DG = x (#29).
Let BE = 3y; then BG = 2y and GE = y (#29).
By the Pythagorean Theorem,
for \triangle DGB, x^2 + (2y)^2 = 9 (#55);
for \triangle EGA, y^2 + (2x)^2 = 16 (#55).
By addition, 5x^2 + 5y^2 = 25;
therefore, x^2 + y^2 = 5.
However, in \triangle BGA, (2y)^2 + (2x)^2 = (AB)^2 (#55),
or 4y^2 + 4x^2 = (AB)^2.
Since x^2 + y^2 = 5, 4x^2 + 4y^2 = 20.
By transitivity, (AB)^2 = 20, and AB = 2\sqrt{5}.
Challenge 1 Express AB in general terms for BC = a, and AC = b.
ANSWER: AB = \sqrt{\frac{a^2 + b^2}{5}}
