Question

If AD is external bisector of $\angle A$ which meets BC at D and CE || DA. Then, which of the following is true?

• A. $\frac{BD}{CD}=\frac{AB}{AC}$
  
• B. $\frac{BD}{CE}=\frac{AB}{AD}$
  
• C. Both (a) and (b)
  
• D. None of the above

Answer 1

The correct option is A

$\frac{BD}{CD}=\frac{AB}{AC}$

Given, AD bisects $\angle A$ externally and CE || AD

$\Rightarrow 2=\angle 4and\angle 3=\angle 1$ [BK, CA respectively being the transversal]
But $\angle 1=\angle 2$ [Since AD bisects $\angle A$ externally]
$\therefore \angle 3=\angle 4$
In $\Delta ACE$,
$\angle 3=\angle 4\Rightarrow AE=AC$ ...(i) [$\because$ sides opposite to equal angles in a triangle are equal]
Now, in $\Delta BAD$, $EC||AD$
$\Rightarrow \frac{BD}{CD}=\frac{BA}{EA}$ [by basic proportionally theorem]
$\Rightarrow \frac{BD}{CD}=\frac{AB}{AC}$ [from Eq. (i)]
Hence option A is the correct option.