# Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC.

Given the triangle ABC whose vertices are endpoints of the altitudes from $$A$$,$$B$$ and $$C$$ is called the orthic triangle. The triangle $$ABC$$ has vertices at $$A=(2,4)$$; $$B=(8,5)$$ and $$C=(3,9)$$. The altitude from $$B$$ to $$AC$$, meets AC at point $$D=(2.42,6.12)$$. Find the area if the orthic triangle.

To attempt this problem I decided to use the formula $$area = \frac{abc|cosAcosBcosC|}{2R}$$ where $$R$$ is the circumradius of the triangle $$ABC$$. I calculated the length of each sides by using the length equation and got $$a=\sqrt41$$, $$b=\sqrt26$$, and $$c=\sqrt37$$.

Next I used the cosine rule to get the angle at vertix $$A$$. Then used the sine rule to get angle at $$B$$ and subtracted these 2 angles from 180 to get the third angle. $$A=69.23$$, $$B=48.12$$ and $$C=62.65$$.

To get the circumradius of the triangle $$ABC$$ I used $$R=\frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks

Find the feet of the altitudes $$D$$, $$E$$, $$F$$ (you already found $$D$$). The area of $$DEF$$ is $$\displaystyle \frac12 \left\Vert \overrightarrow{DE}\times\overrightarrow{DF}\right\Vert$$.
To find $$E$$, we first find the equation of $$(BC)$$: $$\displaystyle y=\dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$$. The equation of the altitude $$(AE)$$ is $$\displaystyle y=5/4 (x-2)+4=5/4 x+3/2$$. To find the intersection $$E$$ of these two lines we solve $$\displaystyle 5/4 x+3/2=-4/5x +11.4$$ we get $$\displaystyle E\left(\frac{198}{41},\frac{309}{41}\right)$$.
After you find $$D$$, $$E$$ and $$F$$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
I suggest to use $$\Delta_{orthic} = 2\Delta\left|\cos A\cos B\cos C\right|$$ and to compute $$\Delta$$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$2\Delta = |2\cdot 5+8\cdot 9+3\cdot 4-4\cdot 8-5\cdot 3-9\cdot 2|=29$$ and by the cosine theorem $$\left|\cos A\cos B\cos C\right|=\frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2}$$ where $$a^2=41,b^2=26,c^2=37$$ are given by the Pythagorean theorem. It follows that $$\Delta_{orthic} = 29\cdot\frac{30\cdot 52\cdot 22 }{8\cdot 41\cdot 26\cdot 37}=\frac{4785}{1517}.$$