Let $\triangle ABC$ an acute triangle and call $K,L, M$ the orthogonal projections of $A,C$ and $B$ on the opposing sides.
Prove: $A_{\triangle KLM} = 2 A_{\triangle ABC}\cdot \cos \hat A \cdot \cos \hat B \cdot \cos \hat C$.
I have tried many approaches but I don't seem to get there. Working with $2A_{\triangle KLM} = |ML|\cdot |KM|\cdot \sin\hat M$ causes issues because I don't see any relationship between $\hat M$ and any other angle.
On the other hand simplyfing the RHS as $|AC|\cdot |CM|\sin \hat C\cos \hat A\cos \hat B$ doesn't seem to work either.
Could anyone give me some pointers on how to tackle the problem?
No full solutions please.
Draw circumcircle of $\triangle ABC$. Extend $AM$ , $BK$ and $CL$ to meet circumcircle of $\triangle ABC$ at $A'$ , $B'$ and $C'$, respectively. Then
Full Solution
Since reflections of orthocenter lies on circumcircle, we have : $HK=KB'$ and $HM=MA'$ so $KM$ is midsegemnt (midline) in triangle $\triangle HB'A'$ which means $KM = \frac{B'A'}{2}$. So we have: $$\frac{KM}{B'A'}=\frac{ML}{A'C'}=\frac{LK}{C'B'}=\frac{1}{2}$$ $\triangle MKL$ and $\triangle A'B'C'$ are similar and $\dfrac{A_{\triangle MKL}}{A_{\triangle A'B'C'}}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$
$A_{\triangle A'B'C'} = \frac{1}{2}.A'B'.A'C'.\sin \angle A'$. Now if we use law of sines we get $$A_{\triangle A'B'C'} = \frac{1}{2}(2R)(2R)\sin \angle A'\sin \angle B'\sin \angle C'$$ So we have to find angles of $\triangle A'B'C'$. $$\angle A' = \angle B'A'A + \angle AA'C'= \angle ACC' + \angle ABB' = 180^\circ -\angle 2A$$ So $$A_{\triangle A'B'C'} = \frac{1}{2}(2R)(2R)\sin \angle 2A \sin \angle 2B\sin \angle 2C$$ Similarly for $\triangle ABC$ we have $A_{\triangle ABC} = \frac{1}{2}(2R)(2R)\sin \angle A \sin \angle B \sin \angle C$ So we have $\dfrac{A_{\triangle A'B'C'}}{A_{\triangle ABC}}=8\cos\angle A \cos\angle B \cos\angle C$