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How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture: bisection

$AQ$ is the bisector of the angle $\angle BAC$, $BR$ -bisector for $\angle ABC$ and $CP$ -bisector for $\angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.

I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.

Thanks :)

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We'll derive the equation using the fact: $$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$ Using the angle bisector theorem we get: $$BP=\frac{ac}{a+b},\quad (1)$$ $$BR=\frac{ac}{b+c}, \quad (2)$$ $$CR=\frac{ab}{b+c},\quad (3)$$ $$CQ=\frac{ab}{a+c},\quad (4)$$ $$AQ=\frac{bc}{a+c},\quad (5)$$ and $$AP=\frac{bc}{a+b}. \quad (6)$$

Each mentioned area can be calculated using:

$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$ $$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$ $$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$ and $$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$

Let $R$ be the circumradius, we know that: $$\sin \alpha = \frac{a}{2R}, \quad (11)$$ $$\sin \beta = \frac{b}{2R}, \quad (12)$$ $$\sin \gamma = \frac{c}{2R}, \quad (13)$$

Now if we substitute all the 13 equations in equation $(I)$ we get: $$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$

$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$

$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$ $$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$ Using Heron's formula we are done.

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This triangle has area $$\frac{2abc}{(a+b)(a+c)(b+c)}\cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."

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