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Math Help - heights and inradius of triangle

  1. #1
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    heights and inradius of triangle

    H_{1}, h_{2}, h_{3} are the heights of a triangle and r is an inradius of this triangle. Prove that:

    \frac{1}{r}= \frac{1}{h_{1} }+\frac{1}{ h_{2} } + \frac{1}{h_{3}}
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  2. #2
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    Hello, bobb12!

    h_a, h_b, h_c \text{ are the altitudes to sides }a,b,c\text{ respectively, of a triangle,}

    . . \text{and }r\text{ is the inradius of this triangle.}

    \displaystyle \text{Prove that: }\;\frac{1}{r} \:=\: \frac{1}{h_1 }+\frac{1}{ h_2 } + \frac{1}{h_3}

    The area \,A of the triangle can be expressed in these four ways:

    . . \begin{array}{ccccccccc}<br />
A \;=\;\frac{1}{2}\,a\!\cdot\!h_a & \Rightarrow & a \:=\:\dfrac{2A}{h_a} &  [1] \\ \\[-3mm]<br />
A \;=\;\frac{1}{2}\,b\!\cdot\!h_b & \Rightarrow & b \:=\:\dfrac{2A}{h_b} &  [2] \\ \\[-3mm]<br />
A \;=\;\frac{1}{2}\,c\!\cdot\!h_c & \Rightarrow & c \;=\;\dfrac{2A}{h_c} & [3] \\ \\[-3mm]<br />
A \;=\;\frac{1}{2}r(a+b+c) & \Rightarrow & \dfrac{2A}{r} \;=\;a+b+c & [4] \end{array}


    \displaystyle\text{Substitute [1], [2], [3] into [4]: }\;\frac{2A}{r} \;=\; \frac{2A}{h_a} + \frac{2A}{h_b} + \frac{2A}{h_c}


    \displaystyle \text{Divide by }2A\!:\;\;\frac{1}{r} \;=\;\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}

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