Seems sticky

Notice that $ABD$ is the result of 'gluing' a 6-8-10 and an 8-15-17 triangle together by the side of length 8, which is the altitude to $BD$ . Similarly, $BCD$ is the 'gluing' of scaled and rotated versions of these triangles; $\frac{45}{4}$ -15- $\frac{75}{4}$ 'glued' with 6- $\frac{45}{4}$ - $\frac{51}{4}$ . Now there are three approaches to find the answer.

• Treating the problem in the plane, with $BD$ on the x-axis, we see that the vertical displacement from $A$ to $C$ is the sum of the altitudes, 8 + $\frac{45}{4}$ = $\frac{77}{4}$ . Each altitude cuts the triangle with lengths of 6, so the horizontal displacement from $A$ to $C$ is 21 - 6 - 6 = 9 = $\frac{36}{4}$ . Here comes another triple! 36-77-85 is a Pythagorean triple so the distance is just $\frac{85}{4}$ .

• Notice that by 'gluing' rotated and scaled right triangles together, we have both triangles $ADC$ and $ABC$ are right-angled, and we have their leg lengths. Note that $ABC$ is a scaled 3-4-5 and that $ADC$ is a scaled 8-15-17 triangle. Cool, right?