Question

In a triangle ABC, let H, I and O be the orthocentre, incentre and circumcentre, respectively. If the points B, H, I, C lie on a circle, what is the magnitude of $\angle$BOC in degree?

Answer 1

In $△ABC$ , $O$ is orthocentre

By property of circle,$\angle BOC=2\angle A$ From fig.$1$

In fig.$2$

In $\square APHQ$

$\angle A+\angle APH+\angle PHQ+\angle HQA={360}^{\circ }$

$\angle HQA=\angle APH={90}^{\circ }$

$⟹\angle A+\angle PHQ={180}^{\circ }$

$\angle PHQ=\angle BHC$ (Vertically opposite angle)

$⟹\angle A+\angle BHC={180}^{\circ }$

$⟹\angle A=-\angle BHC+{180}^{\circ }$

In fig.$3$

$\angle A+\angle B+\angle C={180}^{\circ }$

$\angle AIB=x=\frac{\angle C}{2}$

Similarly, $\angle BIC=y=\frac{\angle B}{2}$

So, $\angle A+2x+2y={180}^{\circ }$

$⟹\frac{\angle A}{2}=90-(x+y)$

In $△CIB$

$\angle BIC+x+y={180}^{\circ }$

$⟹\frac{\angle A}{2}=90-(180-\angle BIC)$

From fig. $4$

$BHIC$ lies on circle

Using property of circle

$\angle BHC=\angle BIC$

$⟹180-\angle A=90+\frac{\angle A}{2}⟹\angle A={60}^{\circ }$

$\angle BOC=2\angle A=2\times {60}^{\circ }={120}^{\circ }$