Question

Length of internal angle bisector drawn through vertex $A$ of triangle $ABC$ whose vertices are $(2,-2),(-1,2)$ and $(3,5)$ respectively is

• A. $\frac{10}{\sqrt{2+\sqrt{2}}}$ units
• B. $\frac{5}{\sqrt{2+\sqrt{2}}}$ units
• C. $\frac{5}{1+\sqrt{2}}$ units
• D. $\frac{10}{1+\sqrt{2}}$ units

Answer 1

The correct option is A $\frac{10}{\sqrt{2+\sqrt{2}}}$ units

$AB=5;BC=5;AC=5\sqrt{2}$


Let internal angle bisector through $A$ meets $BC$ at $D$.
Then $BD:DC=BA:AC=5:5\sqrt{2}=1:\sqrt{2}$
So, $D=(\frac{3-\sqrt{2}}{1+\sqrt{2}},\frac{5+2\sqrt{2}}{1+\sqrt{2}})$
$\Rightarrow AD=\sqrt{{(\frac{3-\sqrt{2}}{1+\sqrt{2}}-2)}^2+{(\frac{5+2\sqrt{2}}{1+\sqrt{2}}+2)}^2}$
$=\frac{\sqrt{100+50\sqrt{2}}}{1+\sqrt{2}}=\frac{10}{\sqrt{2+\sqrt{2}}}$ units