Mathematics - Properties of Triangle Question with Solution | TestHub

We have ${\cos }^2\frac{\mathrm{C}}{2}=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}$ so that $\frac{2\mathrm{ab}}{(\mathrm{a}+\mathrm{b}+\mathrm{c})\Delta }{\cos }^2\frac{\mathrm{C}}{2}=\frac{2\mathrm{ab}}{2\mathrm{s}\Delta }\cdot \frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}=\frac{\mathrm{s}-\mathrm{c}}{\Delta }$

Now, the area of triangle ABC is $\Delta =\frac{1}{2}\mathrm{ap}$, i.e., ${\mathrm{p}}_1=2\Delta /\mathrm{a}$. Similarly, ${\mathrm{p}}_2=2\Delta /\mathrm{b}$ and ${\mathrm{p}}_3=2\Delta /\mathrm{c}$. $\frac{1}{{\mathrm{p}}_1}+\frac{1}{{\mathrm{p}}_2}-\frac{1}{{\mathrm{p}}_3}=\frac{\mathrm{a}+\mathrm{b}-\mathrm{c}}{2\Delta }=\frac{2\mathrm{s}-2\mathrm{c}}{2\Delta }=\frac{\mathrm{s}-\mathrm{c}}{\Delta }$