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The Mid-Point Theorem

In Δ ABC , ...

Question

In $Δ A B C, D E | | B C,$ if $A D = x + 2, D B = 3 x + 16, A E = X a n d E C = 3 x + 5$ then find x.

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Solution

REF.Image.
In $△ A D E$ and $△ A B C$
$∠ A = ∠ A$ [common]
Since $D E ∥ B C,$ DB and EC will
act as traversals
$∴ ∠ A D E = ∠ A B C$
and $∠ A E D = ∠ A C B$ [corresponding angles]
So, by similarity of triangles : $△ A D E \sim △ A B C$ [By AAA citiesion]
$∴ \frac{A D}{A B} = \frac{A E}{A C} \frac{x + 2}{(x + 2) + (3 x + 16)} = \frac{x}{x + (3 x + 5)}$
$\frac{(x + 2)}{(4 x + 18)} = \frac{x}{(4 x + 5)} \Rightarrow (x + 2) (4 x + 5) = x (4 x + 18)$
$\Rightarrow 4 x^{2} + 5 x + 8 x + 10 = 4 x^{2} + 18 x \Rightarrow 5 x = 10 \Rightarrow x = 2$

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