# Area of triangle formed by angle bisector, altitude and median

Question:- Given a triangle ABC with side length a, b and c. Calculate the area of a triangle in terms of a, b and c formed by angle bisector from vertex A, altitude from vertex B and median from vertex C.

looking for solution using complex number, vectors and co-ordinate geometry Regards, vishal

• I hope you find the solution you are looking for. What progress have you made so far? Apr 19, 2015 at 7:58
• my approach for this problem is to get the co-ordinates for the intersections of median, angle bisector and altitude in terms of sides a,b and c and hence calculate the area. But still i am not able to get these co-ordinates in terms of a, b and c. Apr 19, 2015 at 8:30
• Can you get anything, say, the equations of the median, the angle bisector, the altitude? Apr 19, 2015 at 9:46
• You can use Routh's theorem to compute the ratio between the area of your triangle and the area of original triangle. It is relative easy to derive the parameters needed by Routh's theorem and express them in terms of $a, b, c$. After that, you use Heron's formula to express the area of original triangle also in terms of $a,b,c$. Combines these two result, you are done. Apr 19, 2015 at 9:52

Given $\triangle ABC$, $BC=a,\ AC=b,\ AB=c$; $\angle BAD=\angle CAD$, $BE\perp AC$, $AF=BF$, $\angle BAC=\alpha$, $\angle BCA=\gamma$.

First, define coordinates of the points $A,B,C,E$ and $F$: \begin{align} A &=(0,0),\\ B &=(c\cos\alpha,c\sin\alpha), \\ C &=(b,0),\\ E &=(c\cos(\alpha),0),\\ F &=(c/2\cos\alpha,c/2\sin\alpha) \\ \end{align}

Coordinates of the point $D$ can be defined as follows: \begin{align} D_y &= x\tan\gamma = (b-x)\tan\frac{\alpha}{2}, \\ x&=\frac{b}{\tan\gamma+\tan\frac{\alpha}{2}}, \\ D&=\left( \frac{b}{\tan\frac{\alpha}{2}(1+\tan\frac{\alpha}{2}/\tan\gamma)} ,\frac{b}{1+\tan\frac{\alpha}{2}/\tan\gamma} \right). \end{align}

Next, find the intersection points $U=AD \cap CF$, $V=AD \cap BE$, $W=BE \cap CF$:

\begin{align} U&=\left( c b \frac{1+\cos\alpha}{c+2 b} , c b \frac{\sin\alpha}{c+2 b} \right) \\ V &= \left( c\cos\alpha, c\cos\alpha\frac{(1-\cos\alpha)}{\sin\alpha} \right) \\ W &= \left(c \cos(\alpha),c\sin(\alpha) \frac{b-c\cos(\alpha)}{2b-c\cos(\alpha)} \right) \end{align}

\begin{align} S_{\triangle UVW} &= |(V_x-U_x)(W_y-U_y)-(V_y-U_y)(W_x-U_x)|/2. \end{align}

Finally, after substitution $\sin\alpha=\sqrt{1-\cos^2\alpha}$, $\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$ and simplification, the area of $\triangle UVW$ is given in terms of $a,b$ and $c$ by quite a nice construction: \begin{align} S_{\triangle UVW} &= \frac{1}{4} \sqrt{\frac{(a-b+c)(a+b-c)}{(a+b+c)(b+c-a)}} \frac{\left(b^3-b^2 c+b c^2+c^3-a^2 (b+c)\right)^2} {b|3 b^2-c^2+a^2| (c+2 b)} .\quad(*) \end{align}

Test:

the first triangle: $a=8$, $b=10$, $c=12$, $S_{\triangle UVW}=3.266386$,

the second triangle: $a=8$, $b=10$, $c=15$, $S_{\triangle UVW}=16.835510$.

And as expected, when $a=b=c$, $S_{\triangle UVW}=0.$

Edit:

Equation (*) also reveals another extreme case, when the altitude $BE$ is parallel to the median $CF$ and $S_{\triangle UVW}=\infty$ due to the condition $3 b^2-c^2+a^2=0$ in the denominator:

Indeed in this case, $c^2-(2b)^2=a^2-b^2$.