The segment joining the vertices of altitudes A triangle $ABC$ with sides $AB=15,BC=14,AC=13$ is given. $AA_1=12$ and $BB_1$ are heights. Find $A_1B_1$.

I was thinking about the Ptolemy's theorem: $$A_1B_1\cdot CH=B_1C\cdot HA_1+B_1H\cdot CA_1,$$ where $H$ is the intersection of the heights (the orthocenter). We can use it because $HA_1CB_1$ is cyclic: $$\measuredangle HB_1C+\measuredangle HA_1C=180^\circ.$$ I was able to find that $CA_1=5$. I am not sure this is the easiest solution, because we have to find $5$ more segments. Any thoughts on the problem will be appreciated. Thank you!

Equate the area of the triangle to find $BB_1 = \frac{12 \times 14}{13}$.
$A_1C = 5$ as you mentioned.
Applying Pythagoras, $B_1C^2 = BC^2 - BB_1^2 \implies B_1C = \frac{70}{13}$
Now find $\cos C = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} = \frac{5}{13}$
$A_1B_1^2 = A_1C^2 + B_1C^2 - 2 A_1C \ B_1C \cos C = 25 + (\frac{70}{13})^2 - \frac{3500}{13^2} = (\frac{75}{13})^2$
Hence $A_1B_1 = \frac{75}{13}$
