Given the area, perimeter and an altitude of a triangle, find some products The triangle $ABC$ has perimeter $360$ and area $2100$. The altitudes are named $AD$, $BE$, $CF$.
The altitudes meet at $H$. If $CF=24$, find the values of $HA \times HD$, $HB \times HE$, $HC \times HF$.
I have observed that all of those products are equal (because of some similar triangles), but I don't know how to find them. Can you help me? Thanks!
 A: Draw circles $(ABDE), (BECF), (CFAD)$. It is clear that $H$ is the radical center of all three circles. Thus, $HA \cdot HD = HB \cdot HE = HC \cdot HF = x$.
Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\cdot CF = 4200$. It follows that
$$
\begin{align}
\sqrt{a^2 - 24^2} + \sqrt{b^2 - 24} = AB &= 175 \\
a+b+175 &= 360 
\end{align}
$$
Thus, examining Pythagorean triples, $a = 40$, $b = 145$. By the Pythagorean Theorem, we can establish that $AF = 143$, and from the given area, $AD = 105$, so $DC = 100$ and $BD = 140$.
Finally, $\triangle AFH \sim \triangle CDH \sim ABD$, so
$$x = HC \cdot HF = \frac{BD}{AD} \cdot \frac{AB}{AD} \cdot AF \cdot DC = \frac{140}{105} \cdot \frac{175}{105} \cdot 143 \cdot 100 = \boxed{\frac{286000}{9}}$$
A: Hint:  If you know the perimeter, area, and an altitude, then you can solve for the side lengths of the triangle.
To do this express the area in two ways.  First $$Area = \frac{1}{2} \text{base} \times \text{height}$$ which gives us one base length (in your case use $CF$ as the height to isolate $AB$).
Secondly use Heron's formula for the area (with $P$ the perimeter):
$$Area = \sqrt{\frac{P}{2} \left(\frac{P}{2} - AB\right)\left(\frac{P}{2} - BC\right)\left(\frac{P}{2} - AC\right)}$$
in conjunction with $P = AB + BC + AC$ to solve for the remaining lengths.  
Knowing all of the side lengths and the altitude $CF$ you can now solve for $BF$ and $AF$, using the pythagorean theorem.  Finally by considering similar triangles you are in a position to finish the problem.  
