Question

The internal angular bisectors of angles of a $\Delta ABC$ meets circumcircle of $\Delta ABC$ at $D,E,F$ respectively then $\frac{\text{Area of}{\Delta }^{le}\text{ABC}}{\text{Area of}{\Delta }^{le}\text{DEF}}=$

• A. $\frac{r}{R}$
• B. $\frac{2r}{R}$
• C. $\frac{3r}{R}$
• D. $\frac{4r}{R}$

Answer 1

The correct option is B $\frac{2r}{R}$

We know, angle on same chord subtends equal angles on the circumference. So, we have:
$\angle BED=\angle BAD=\frac{A}{2}$
Similarly, $\angle FEB=\angle FCB=\frac{C}{2}$
$\therefore \angle FED=\frac{A+C}{2}={90}^0-\frac{B}{2}$
Thus we have:
$\angle E={90}^0-\frac{B}{2}$
and Similarly, $\angle D={90}^0-\frac{A}{2}$
$\angle F={90}^0-\frac{C}{2}$
We know, area of ${\Delta }^{le}\text{ABC}=\frac{abc}{4R}$
Now, since circumradius is same for both the triangles, the ratios of the areas can be written as the ratios of sine of corresponding sides.
$\therefore \frac{\text{Area of}{\Delta }^{le}\text{ABC}}{\text{Area of}{\Delta }^{le}\text{DEF}}=\frac{\sin A\sin B\sin C}{\sin \left({90}^0-\frac{A}{2}\right)\sin \left({90}^0-\frac{B}{2}\right)\sin \left({90}^0-\frac{C}{2}\right)}$
$\frac{\text{Area of}{\Delta }^{le}\text{ABC}}{\text{Area of}{\Delta }^{le}\text{DEF}}=\frac{\sin A\sin B\sin C}{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}$
$\Rightarrow \frac{\text{Area of}{\Delta }^{le}\text{ABC}}{\text{Area of}{\Delta }^{le}\text{DEF}}=8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{B}{2}=\frac{2r}{R}$