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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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2 Answers 2

up vote 1 down vote accepted

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".

A reference: http://mathworld.wolfram.com/IncentralTriangle.html

This other reference is to Kimberling's triangle center page, a huge thing describing a vast number of triangle centers such as the incenter, and in many cases giving the area of the cevian triangle formed as you did by drawing the lines through the center from the vertices of the reference triangle to the opposite sides. The reference is http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

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