# Area of triangle formed by the feet of the altitudes.

I have this question I can't seem to find a solution to:

$\triangle$ABC is a triangle such that AD is the altitude from A to BC, BE is the altitude from B to AC and CF is the altitude from C to AB.

Now complete $\triangle$DEF.

Find the ratio of Area $\triangle$DEF to Area $\triangle$ABC.

I think it has to something to do with the incenter- excenter lemma( Although I couldn't solve it that way). I tried using the sine rule and the fact that the line OD, OE and OF bisect the interior angles of $\triangle$DEF. And I'm stuck.

The answer is in terms of $\sin$ and cos of the angles A, B, C according to the answer key.

Any and all help is appreciated!

To expand on my comment, the trick is that if the original triangle has circumradius $R$ then the orthic triangle has circumradius $R/2$ because its vertices lie on the celebrated nine-point circle, so one of its side lengths is $R|\sin (\pi-2A)|=a|\cos A|$. Its area is therefore $$\frac{1}{2}a|\cos A|\cdot b|\cos B|\cdot |\sin (\pi-2C)|=\frac{1}{2}ab\sin C\cdot \color{blue}{2|\cos A\cos B\cos C|}.$$The factor in blue is the desired area ratio. As a sanity check, for an equilateral triangle it's $2\cdot(\frac{1}{2})^3=\frac{1}{4}$, which is what we would expect.