Question

If $P_1,P_2,P_3$ are the altitudes of a triangle from the vertices $A,B,C$ & $\Delta$ denotes the area of triangle, then $\frac{1}{P_1}+\frac{1}{P_2}-\frac{1}{P_3}=\frac{2ab}{(a+b+c)\Delta }co{s}^2\frac{C}{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

Since $\Delta =\frac{1}{2}a{p}_1\Rightarrow \frac{1}{p_1}=\frac{a}{2\Delta }$

$\Delta =\frac{1}{2}b{p}_2\Rightarrow \frac{1}{p_2}=\frac{b}{2\Delta }$

$\Delta =\frac{1}{2}c{p}_3\Rightarrow \frac{1}{p_3}=\frac{c}{2\Delta }$

$\therefore \frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}=\frac{1}{2\Delta }(a+b-c)$

$=\frac{2(s-c)}{2\Delta }=\frac{s-c}{\Delta }=\frac{s(s-c)}{ab}.\frac{ab}{s\Delta }$

$=\frac{ab}{\left(\frac{a+b+c}{2}\right)\Delta }{\cos }^2\frac{C}{2}$

$=\frac{2ab}{(a+b+c)\Delta }{\cos }^2\frac{C}{2}$

Hence the given statement is true.