Mathematics - Properties of Triangle Question with Solution | TestHub

Let $ABC$ be the isosceles triangle with $\mathrm{AB}=\mathrm{AC}=\mathrm{b}$ and $\angle \mathrm{B}=\angle \mathrm{C}=\alpha$. Let $AD$ be the perpendicular bisector of the side $BC$. Since $△ABC$ is isosceles, $AD$ is also the bisector of angle $A$. So that $O$ and $I$ both lie on $AD$. We have $OB=R$ and $ID=r$. Also, since $O$ is the circumcenter, we get $OA=OB=R$. Therefore, from isosceles triangle $OAB$:

$$\frac{\mathrm{OB}}{\sin \left(90^{\circ }-\alpha \right)}=\frac{\mathrm{AB}}{\sin 2\alpha }\Rightarrow \mathrm{R}=\frac{\mathrm{b}\cos \alpha }{2\sin \alpha \cos \alpha }$$

$=\frac{1}{2}\mathrm{b}\mathrm{cosec}\alpha$

(a) is correct. Again, $\Delta =\mathrm{BD}\cdot \mathrm{AD}=\mathrm{b}\cos \alpha \cdot \mathrm{b}\sin \alpha =\frac{1}{2}{\mathrm{b}}^2\sin 2\alpha$.

$r=\frac{\Delta }{s}=\frac{\frac{1}{2}{b}^2\sin 2\alpha }{\frac{1}{2}(b+b+2b\cos \alpha )}=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$

(c) is correct. Further, $\mathrm{OI}=|\mathrm{OD}+\mathrm{DI}|=|\mathrm{OD}+\mathrm{r}|$. Because $\alpha <\frac{\pi }{4}$, $\mathrm{A}>\frac{\pi }{2}$ and $O$ lies on $AD$ produced. Now, from right-angled triangle $ODB$, we get ${\mathrm{OD}}^2={\mathrm{OB}}^2-{\mathrm{BD}}^2={\mathrm{R}}^2-(\mathrm{b}\cos \alpha {)}^2$

$=\frac{1}{4}\frac{b^2}{{\sin }^2\alpha }-b^2{\cos }^2\alpha$

$=\frac{b^2\left(1-4{\sin }^2\alpha {\cos }^2\alpha \right)}{4{\sin }^2\alpha }$ [from (a)]

$=\frac{b^2{\left({\cos }^2\alpha -{\sin }^2\alpha \right)}^2}{4{\sin }^2\alpha }=\frac{b^2{\cos }^{2}2\alpha }{(2\sin \alpha {)}^2}$

$\Rightarrow \mathrm{OD}=\frac{\mathrm{b}\cos 2\alpha }{2\sin \alpha }$

$\therefore \mathrm{OI}=\left|\frac{\mathrm{b}\sin 2\alpha }{2(1+\cos \alpha )}+\frac{\mathrm{b}\cos 2\alpha }{2\sin \alpha }\right|$

$=\left|\frac{b\sin 2\alpha }{4{\cos }^2(\alpha /2)}+\frac{b\cos 2\alpha }{4\sin (\alpha /2)\cos (\alpha /2)}\right|$

$=\left|\frac{b}{4\cos (\alpha /2)}\cdot \left(\frac{\sin 2\alpha \sin (\alpha /2)+\cos 2\alpha \cos (\alpha /2)}{\sin (\alpha /2)\cos (\alpha /2)}\right)\right|$

$=\frac{\mathrm{b}\cos (3\alpha /2)}{2\sin \alpha \cos (\alpha /2)}$ (d) is also correct