In ΔABC the altitude, angle bisector, and median from C divide ∠C into four equal angles.

What is the ratio of sides a and b (shorter:longer)?

Trying to avoid the trig....

Let points D, E and F, on AB, lie on the altitude, bisector and median from C.
CE = b since triangles ACD and ECD are congruent.

CE bisects /ACB, so        |BE| = ac/(a + b)

CF bisects /BCE, so         |BF| = a|BE|/(a + b) = a²c/(a + b)²

|BF| = c/2, therefore      a²c/(a + b)² = c/2

which gives        (a + b)/a = sqrt(2)         and therefore:   b/a = sqrt(2) - 1

Posted by Harry on 2010-12-03 13:43:43