$\Delta ABC$ has heights $AD,BE,CF$ and circumradius R, prove $(DEF)=\frac12R^2\sin2A\sin2B\sin2C$ If the acute triangle $\Delta ABC$ has heights $AD,BE,CF$ and circumradius R, prove the following formulas for the area:
$(ABC)=2R^2\sin A\sin B\sin C$
$(DEF)=\frac12R^2\sin2A\sin2B\sin2C$
With the first one I had no trouble, I used the Law of sines and that$(ABC)=\frac12ab\sin C$. I am having trouble with the second one
The only thing that I can think of is that $\Delta AOC, \Delta AOB, \Delta BOC$ are isosceles, with $O$ being the centre of the triangle.
Can I get some help?
 A: $$\measuredangle FED=\measuredangle FEB+\measuredangle DEB=\measuredangle FAD+\measuredangle DCF=90^{\circ}-\beta+90^{\circ}-\beta=180^{\circ}-2\beta.$$
Also, since $$\Delta AFE\sim\Delta ACB,$$ we obtain:
$$\frac{FE}{BC}=\frac{AF}{AC}=\cos\alpha,$$ which gives $$FE=BC\cos\alpha=2R\sin\alpha\cos\alpha=R\sin2\alpha.$$
Similarly, $$DE=R\sin2\gamma.$$
Id est, $$S_{\Delta FED}=\frac{1}{2}R\sin2\alpha\cdot  R\sin2\gamma\cdot\sin(180^{\circ}-2\beta)=$$
$$=\frac{1}{2}R^2\sin2\alpha\sin2\beta\sin2\gamma.$$
A: 
We have, $[DEF]=[ABC]−[AEF]−[BFD]−[CDE]$

Consider the $\triangle AEF$, $$[AEF]=\frac{1}{2}AQ\cdot AR\cdot\sin A.$$
But we know that $[ABC]=\frac{1}{2}b\cdot c\cdot\sin A,$ and so $\sin A=2\frac{[ABC]}{bc}.$
Substituting, we get $$[AEF]=\frac{AE\cdot AF}{b\cdot c}[ABC].$$
But if we look at the right-angled triangles $FAC$ and $BAE$, we see that $$\cos A=\frac{AF}{b}=\frac{AE}{c},$$
and substituting gives $$[AQR]=\cos^2A[ABC].$$

Similarly, $[BDF]=\cos^2B[ABC]$
and $[CED]=\cos^2C[ABC].$
Finally, $$[DEF]=[ABC]−[AEF]−[BDF]−[CDE]=(1−\cos^2A−\cos^2B−\cos^2C)[ABC].$$
Now, you just have to prove that $$1−\cos^2A−\cos^2B−\cos^2C=2\cos A\cdot\cos B\cdot\cos C$$ You can check it here.
