A Strange Cevian

I used analytic geometry to help me here. Place point $B$ on the origin with $\overline{BC}$ on the x-axis. Then $C = (14,0)$ .

By the Law of Cosines, $\cos(\angle ABC) = \dfrac{13^2+14^2-15^2}{2*13*14} = \dfrac{5}{13}$ . Therefore, $\sin(\angle ABC) = \dfrac{12}{13}$ , and thus $A = (5, 12)$ .

The semiperimeter, $s$ , of the triangle is $\dfrac{13+14+15}{2} = 21$ . The formula for the radius of the incircle is: $r = \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}} = \sqrt{\dfrac{(21-13)(21-14)(21-15)}{21}} = 4$

Let $E, F, G$ be the points of tangency of the circle to the sides $\overline{AB}, \overline{BC}, \overline{CA}$ respectively. Let $p = \overline{AE} = \overline{AG}; q = \overline{BE} = \overline{BF}; r = \overline{CF} = \overline{CG}$ . Then we have this system of equations:

$\left\{\begin{array}{lr} p + q = 13 \\ q + r = 14 \\ r + p = 15 \end{array} \right.$

Solving this, we get $p = 7, q = 6, r = 8$ . Therefore, $F = (6, 0)$ .

Therefore, the centre of the circle is $(6, 4)$ , so its equation is $(x - 6)^2 + (y - 4)^2 = 16$ .

Since $AB + BD = AC + CD$ , and since $BD + CD = 14$ , we get that $\overline{BD} = 8$ and $\overline{CD} = 6$ . Therefore $D = (8, 0)$ .

Therefore, the equation of the line $\overline{AD}$ is: $\dfrac{y - 12}{0 - 12} = \dfrac{x - 5}{8 - 5}$ $y = -4x + 32$

Now we just need to find the intersection point of the circle and the line:

$(x - 6)^2 + (-4x + 32 - 4)^2 = 16$ $x^2 - 12x + 36 + 16x^2 - 224x + 784 = 16$ $17x^2 - 236x + 804 = 0$ $(17x - 134)(x - 6) = 0$ Thus $x = 6$ or $x = \dfrac{134}{17}$ . Thus, the two possibilities for $X$ are $(6, 8)$ and $\left(\dfrac{134}{17}, \dfrac{9}{17}\right)$ . The former is clearly closer to $A$ .

Thus, $\overline{BX} = \sqrt{6^2+8^2} = \boxed{10}$ .

Point $D$ is actually the tangency point of the $A$ -excircle with $BC.$ Also, $X$ is the "top" part of the incircle i.e. the diametrically opposite point from the tangency point of the incircle with $BC.$ Let $Y$ be the point of tangency of $BC$ with the incircle. It is clear that $BXY$ is a right triangle. We have $XY = 2r = 8$ and $BY = s - b = 21 - 15 = 6,$ so from the Pythagorean Theorem, $BX = \boxed{10}.$

Can this be drawn in Geogebra to verify that BX is exactly 10? My answer was close to 10.1 using Stewart's Theorem for Tr. ABD which would suggest that X is not exactly opposite the point of tangency on BC.