Properties of Triangles 2 Question 15
15. If $p _1, p _2, p _3$ are the altitudes of a triangle from the vertices $A, B, C$ and $\Delta$ is the area of the triangle, then prove that
$$ \frac{1}{p _1}+\frac{1}{p _2}-\frac{1}{p _3}=\frac{2 a b}{(a+b+c) \Delta} \cos ^{2} \frac{C}{2} $$
(1978, 3M)
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Solution:
- Since, $\Delta=\frac{1}{2} a p _1 \Rightarrow \frac{1}{p _1}=\frac{a}{2 \Delta}$
Similarly, $\frac{1}{p _2}=\frac{b}{2 \Delta}, \frac{1}{p _3}=\frac{c}{2 \Delta}$
$$ \begin{aligned} \therefore \quad \frac{1}{p _1}+\frac{1}{p _2}-\frac{1}{p _3} & =\frac{1}{2 \Delta}(a+b-c) \\ & =\frac{2(s-c)}{2 \Delta}=\frac{s-c}{\Delta}=\frac{s(s-c)}{a b} \cdot \frac{a b}{s \Delta} \\ & =\frac{a b}{\frac{a+b+c}{2} \Delta} \cdot \cos ^{2} \frac{C}{2} \\ & =\frac{2 a b}{(a+b+c) \Delta} \cos ^{2} \frac{C}{2} \end{aligned} $$
