Let D be a point on side BC of triangle ABC.

If the incircles of triangles ABD and ACD are congruent, then what is the length of the cevian AD in terms of a, b, and c ?

(In reply to question by Dej Mar)

The answer is yes (if we can say the following is easy):

Given AE perpendicular to AB at point E we have a segment of a length equal to the height of ABC. We know DC = DE + CE, thus DE = DC - CE. Using the Pythagorean theorem we have
cevian AD = SQRT(DE² + AE²) and CE = SQRT(b² - AE²), thus cevian AD = SQRT((a - SQRT(b² - AE²) + AE²).

Using the conventional method of calculating the area of triangle ABC we have the area of triangle ABC 
= 1/2*a*AE.
Using Heron's formula, the area of triangle ABC
= SQRT(s(s-a)(s-b)(s-c)),
such that s is the semiperimeter of triangle ABC, i.e.,
s = (a+b+c)/2.

Using the two equations we have
AE = 2/a * (SQRT(s*(s - a)*(s - b)*(s - c)))
Thus, AE² = 4/a² * (s*(s - a)*(s - b)*(s - c))

By substitution we have cevian AD in terms of a, b and c, i.e.,
AD = SQRT((a - SQRT(b² - 4/a² * ((a+b+c)/2 * ((a+b+c)/2 - a)* ((a+b+c)/2 - b) * ((a+b+c)/2 - c)) + 4/a² * ((a+b+c)/2 * ((a+b+c)/2 - a) * ((a+b+c)/2 - b) * ((a+b+c)/2 - c))

Edited on May 16, 2008, 1:00 pm

Posted by Dej Mar on 2008-05-16 02:31:55