Question

In triangle ABC, side AC and the perpendicular bisector of side BC meet at point D and BD bisects $\angle ABC.$ If AD = 9 AND DC = 7.Find area of $\Delta$ ABD

• A.
  
• B. 25
• C.
  
• D.
  

Answer 1

The correct option is C

By angle bisector theorem $\frac{AB}{BC}=\frac{9}{7}$
Let AB = 9x and BC = 7x
We have BD = CD = 7 (D equidistant from B and C)
Also $\angle ADB$ is exterior angle of $\Delta$BDC
$\angle ADB=\angle DBC+\angle DCB=2\angle DBC=\angle ABC$
$\angle BAC$ is common, therefore $\Delta ADB\approx \Delta ABC\frac{AD}{AB}=\frac{AB}{AC}\left(CPCT\right)=\frac{9}{9x}=\frac{9x}{16}X=\frac{4}{3}\therefore AB=12$
Area of $\Delta ABD=14\sqrt{5.}$