Question

If $A(8,12),B(8,0)$ and $C(0,6)$ are the vertices of a triangle $ABC$, then

• A. Length of angle bisector through vertex $B$ is $\frac{48}{11}\sqrt{5}$ units
• B. Length of angle bisector through vertex $B$ is $\frac{24}{11}\sqrt{5}$ units
• C. Incentre divides the angle bisector through $B$ in $11:5$
• D. Incentre divides the angle bisector through $B$ in $11:9$

Answer 1

Answer: (A), (C)

Incentre divides the angle bisector through $B$ in $11:5$

For triangle $ABC,AB=12,BC=10,AC=10$



As angular bisector through a vertex divides opposite side in the ratio of it sides i.e.,$\frac{CD}{DA}=\frac{BC}{AB}=\frac{10}{12}=\frac{5}{6}$
$\Rightarrow \frac{CD}{DA}=\frac{5}{6}$

$\Rightarrow D\equiv (\frac{40+0}{5+6},\frac{60+36}{5+6})$
$\Rightarrow D\equiv (\frac{40}{11},\frac{96}{11})$

Now, $BD=\sqrt{{(\frac{40}{11}-8)}^2+{(\frac{96}{11}-0)}^2}$
$BD=\frac{48}{11}\sqrt{5}$ units

Since, Incentre divides the angular bisector through vertex $B$ in the ratio $a+c:b$
$\therefore \frac{BI}{ID}=\frac{12+10}{10}=\frac{11}{5}$