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!
 A: 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 $A_1B_1 = \frac{75}{13}$
A: The area of the triangle is, by Heron's formula, $$|\triangle ABC| = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(6)(7)(8)} = 84,$$ thus $$AA_1 = \frac{2(84)}{14} = 12, \quad BB_1 = \frac{2(84)}{13} = \frac{168}{13},$$ and by the Pythagorean theorem, $$CA_1 = 5, \quad CB_1 = \frac{70}{13}.$$  Since by the Law of Cosines we have $$\cos \angle C = \frac{13^2 + 14^2 - 15^2}{2(13)(14)} = \frac{5}{13},$$ we obtain
$$(A_1 B_1)^2 = (CA_1)^2 + (CB_1)^2 - 2(CA_1)(CB_1) \cos \angle C = \left(\frac{75}{13}\right)^2,$$ and the desired length is $75/13$.
A: Note that $ABA_1B_1$ is cyclic, implying similar triangles $ABC $ and $A_1B_1C$. Thus,
$$ A_1B_1= \frac{CA_1}{CA}AB =\frac 5{13}\cdot15= \frac{75}{13}$$
