# heights and inradius of triangle

• Jan 29th 2011, 03:28 AM
bobb12
$H_{1}, h_{2}, h_{3}$ are the heights of a triangle and r is an inradius of this triangle. Prove that:

$\frac{1}{r}= \frac{1}{h_{1} }+\frac{1}{ h_{2} } + \frac{1}{h_{3}}$
• Jan 29th 2011, 06:06 AM
Soroban
Hello, bobb12!

Quote:

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

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

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

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

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

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

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