Question

In a $∆$ABC, the sides AB and AC are produced to the points P and Q respectively. The bisectors of $\angle$PBC and $\angle$QCB intersect at a point O. prove that $\angle$BOC = 90 $-\frac{1}{2}$ $\angle$A

Answer 1

$$\begin{gathered} \mathrm{Let}\angle \mathrm{PBO}=\angle \mathrm{OBC}=y\mathrm{and}\angle \mathrm{BCO}=\angle \mathrm{OCQ}=\mathit{}x \\ \mathrm{The}\mathrm{angles}\angle \mathrm{PBC}\mathrm{and}\angle \mathrm{QCB}\mathrm{are}\mathrm{bisected}. \\ \mathrm{Then}\mathrm{in}△\mathrm{BOC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BOC}+x+y=180° \\ \Rightarrow \angle \mathrm{BOC}=180°-x-y...\left(\mathrm{i}\right) \\ \mathrm{Also},\angle \mathrm{ABC}+2y=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow \angle \mathrm{ABC}=180°-2y...\left(\mathrm{ii}\right) \\ \mathrm{and}\angle \mathrm{ACB}+2x=180° \\ \Rightarrow \angle \mathrm{ACB}=180°-2x.....\left(\mathrm{iii}\right) \\ \mathrm{Similarly},\mathrm{in}∆\mathrm{ABC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180° \\ \Rightarrow \angle \mathrm{BAC}+180°-2y+180°-2x=180°[\mathrm{From}(\mathrm{ii})\mathrm{and}(\mathrm{iii}\left)\right] \\ \Rightarrow \angle \mathrm{BAC}=2x+2y-180°....\left(\mathrm{iv}\right) \\ \mathrm{Now},\mathrm{RHS}=90°-\frac{1}{2}\angle \mathrm{A} \\ =90°-\frac{1}{2}\left(2x+2y-180°\right) \\ =90°-x-y+90° \\ =180°-x-y \\ =\angle \mathrm{BOC} \\ =\hspace{0.17em}\mathrm{LHS} \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$