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Internal Angle Bisector Theorem

DE is drawn p...

Question

$D E$ is drawn parallel to the base $B C$ of a $△ A B C$ , meeting $A B$ at $D$ and $A C$ at $E, \frac{A B}{A C} = 4$ and $C E = 2 c m$ , find $A E$ .

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Solution

$\begin{matrix} g i v e n \frac{A D}{D B} = 4, E C = 2 C M w e k n o w t h a t \frac{A D}{D B} = \frac{A E}{E C} 4 = \frac{A E}{2} ∴ A E = 8 c m \end{matrix}$

Hence, this is the answer.

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