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)

Show Answer

Solution:

  1. Since, $\Delta=\frac{1}{2} a p _1 \Rightarrow \frac{1}{p _1}=\frac{a}{2 \Delta}$

Similarly, $\frac{1}{p _2}=\frac{b}{2 \Delta}, \frac{1}{p _3}=\frac{c}{2 \Delta}$

$$ \begin{aligned} \therefore \quad \frac{1}{p _1}+\frac{1}{p _2}-\frac{1}{p _3} & =\frac{1}{2 \Delta}(a+b-c) \\ & =\frac{2(s-c)}{2 \Delta}=\frac{s-c}{\Delta}=\frac{s(s-c)}{a b} \cdot \frac{a b}{s \Delta} \\ & =\frac{a b}{\frac{a+b+c}{2} \Delta} \cdot \cos ^{2} \frac{C}{2} \\ & =\frac{2 a b}{(a+b+c) \Delta} \cos ^{2} \frac{C}{2} \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane