Question

If $p_1,p_2,p_3$ are the altitudes of a triangle from the vertices A, B, C and $△$ , the area of the triangle, prove that $\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}=\frac{2ab}{(a+b+c)△}{\cos }^2\frac{C}{2}$.

Answer 1

Since $p_1,p_2,p_3$ are perpendiculars from the vertices A, B, C to the opposite sides, we have
$△=\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{P}_3$
Hence $\frac{1}{p_1}+\frac{1}{p_2}\frac{1}{p_3}=\frac{a}{2△}+\frac{b}{2△}-\frac{c}{2△}$
=$\frac{a+b-c}{2△}=\frac{a+b+c-2c}{2△}=\frac{2s-2c}{2△}$
$=\frac{s-c}{△}=\frac{ab}{△s}.\frac{s(s-c)}{ab}$
=$\frac{ab}{△s}{\cos }^2\frac{1}{2}C=\frac{1}{2}C=\frac{2ab}{(a+b+c)△}{\cos }^2\frac{1}{2}C.$