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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.

Diagram

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!

Thanks in advance.

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  • $\begingroup$ The title should summarize the question (and "trigonometry / geometry question" isn't really enough, even though it's pretty common around here), not be a plea for help. $\endgroup$ – Arthur Sep 4 '18 at 15:09
  • $\begingroup$ ok fine i'll change it. Thanks for the edits. $\endgroup$ – ujwal kumar Sep 4 '18 at 15:11
  • $\begingroup$ Welcome to Math.SE! You should include your thoughts about the problem at hand, and give an idea of where exactly you got stuck. This will help potential answers tailor their responses to your skill level ... and it will help convince them that you aren't simply trying to get them to do your homework for you. $\endgroup$ – Blue Sep 4 '18 at 15:14
  • $\begingroup$ That's the orthic triangle. Proposition 3 proven in this link gives its area: irmo.ie/5.Orthic_triangle.pdf Get the rest using the extended sine rule. $\endgroup$ – J.G. Sep 4 '18 at 15:24
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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.

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