Question

In triangle ABC, BD is the perpendicular bisector of BC meets AC in point D the BD is angle bisector of angle ABC if AD=9 DC=7 then find the area of triangle ABD.

Answer 1

$\Delta BDC$, D is vertex and lies on $\perp$ trisector
$\therefore \Delta BDC$ is isoceles $\Rightarrow$ BD = DC = 7
Let $C=x\Rightarrow \angle DBC=x$
$\Rightarrow \angle ABD=x$ (since BD is angular bisector)
$\therefore \angle ADB=2x$
$\Delta ABD$ and $\Delta ACB$ are similar
$\because \angle ADB=\angle ABC=2x$
$\angle DBA=\angle BCD=\angle BCA=x$
$\therefore \frac{AB}{BC}=\frac{AD}{AB}$
$\frac{AB}{16}=\frac{9}{AB}\Rightarrow A{B}^2=16\times 9$
or $AB=4\times 3=12$
$\therefore$ Area of $\Delta$ ABD using Heron's formula
(sides 12,7,9)
$=\sqrt{\frac{28}{2}(14-12)(14-7)(14-9)}=\sqrt{14\times 2\times 7\times 5}=14\sqrt{5}sq.units$