<s>[INST] Geometry problem about triangle orthocenter Let $ABC$ be a triangle with orthocenter $H$. Let $AB = 84$, $AC = 32\sqrt{3}$ and $\angle{BAC} = 60^\circ$. Let $D, E, F$ be points on $AH, BH, CH$,    respectively, such that: $7AD = AH$, $7BE = BH$, $7CF = CH$. Find $AB\cdot CF + BC\cdot AD + AC\cdot BE$.
This problem is taken from a math competition and you should be able to solve it without the aid of a calculator; I managed to find both $CF$ and $BE$, using similar triangles. However I'm stuck with $AD$: the expression that leads to the evaluation of $AD$ is simply too difficult (a lot of square roots), so I was wondering if someone could give me a hint or sketch the solution. Any help would be greatly appreciated.
 [/INST] On each point $A, B, C$ of the triangle, pass through it a line parallel to its opposite side. Then, these 3 new lines form a new triangle, similar to the original one. the orthocenter of $ABC$ coincide with the circumcenter of the new triangle. Notice that
$$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = AB \cdot CH + BC \cdot AH + AC \cdot BH$$
$$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = \frac{2AB \cdot CH}{2} + \frac{2BC \cdot AH}{2} + \frac{2AC \cdot BH}{2}$$
Now, the sides of the new triangle are given by $2AB$, $2BC$ and $2CA$, so the expression above corresponds to its area. Thus, we have
$$7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = 4 \text{Area}(ABC) = 4 \frac{84 \cdot 32\sqrt{3} \sin 60^\circ}{2} = 2 \cdot 84 \cdot 32 \cdot \frac{3}{2}$$
implying that
$$ AB\cdot CF + BC\cdot AD + AC\cdot BE = 12 \cdot 32 \cdot 3 = 9 \cdot 128 = 1152.$$
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