In given figure, XY||AC and XY divides triangular region ABC into two parts equal in area. Determine $\frac{A X}{A B}$ .

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Solution

We have XY||AC

and, $A r e a (△ B X Y) = A r e a (q u a d X Y C A)$

$\Rightarrow$ $A r e a (△ A B C) = 2 A r e a (△ B C Y)$ ........(i)

Now, $X Y | | A C$ and BA is a transversal.

$\Rightarrow$ $∠ B X Y = ∠ B A C$ .......(ii)

Thus, in $△^{'} s B A C$ and $B X Y$ , we have

$∠ X B Y = ∠ A B C$ [Common]

$∠ B X Y = ∠ B A C$ [From(ii)]

Therefore, AA-criterion of similarity, we have

$△ B A C \sim △ B X Y$

$\Rightarrow$ $\frac{A r e a (△ B A C)}{A r e a (△ B X Y)} = \frac{{B A}^{2}}{{B X}^{2}}$

$\Rightarrow$ $2 = \frac{{B A}^{2}}{{B X}^{2}}$ [Using (i)]

$\Rightarrow$ $B A = \sqrt{2} B X \Rightarrow B A = \sqrt{2} (B A - A X)$

$\Rightarrow$ $(\sqrt{2} - 1) B A = \sqrt{2} A X$

$\Rightarrow$ $\frac{A X}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

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