Three forces P, Q and R acting along IA, IB and IC, where I is the incentre of a ABC, are …

Question

Three forces $\vec P,\vec Q$ and $\vec R$ acting along $IA$, $IB$ and $IC$, where $I$ is the incentre of a $	riangle ABC$, are in equilibrium. Then $\vec P:\vec Q:\vec R$ is:

Accepted Answer

Since $IA,IB,IC$ are angle bisectors, $\angle BIC=90^{\circ}+\dfrac A2$, and similarly for the others. By Lami's theorem at $I$:

$$\frac{P}{\sin\left(90^{\circ}+\frac A2\right)}=\frac{Q}{\sin\left(90^{\circ}+\frac B2\right)}=\frac{R}{\sin\left(90^{\circ}+\frac C2\right)} \implies P:Q:R=\cos\frac A2:\cos\frac B2:\cos\frac C2$$

Solutions · 1