In $△$ ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ $∥$ BC and PQ divides $△$ ABC into two parts equal in area. Find $\frac{B P}{A B}$ .

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Solution

Since the line $P Q$ divides $△ A B C$ into two equal parts,

$a r e a (△ A P Q) = a r e a (△ B P Q C)$

$\Rightarrow$ $a r e a (△ A P Q) = a r e a (△ A B C) - a r e a (△ A P Q)$

$\Rightarrow$ $2 a r e a (△ A P Q) = a r e a (△ A B C)$

$∴$ $\frac{a r e a (△ A B C)}{a r e a (△ A P Q)} = \frac{2}{1}$ ---- ( 1 )

Now, in $△ A B C$ and $△ A P Q$ ,

$∠ B A C = ∠ P A Q$ [ Common angles ]

$∠ A B C = ∠ A P Q$ [ Corresponding angles ]

$∴$ $△ A B C \sim △ A P Q$ [ By AA similarity ]

$∴$ $\frac{a r e a (△ A B C)}{a r e a (△ A P Q)} = \frac{A B^{2}}{A P^{2}}$

$∴$ $\frac{2}{1} = \frac{A B^{2}}{A P^{2}}$ [ From ( 1 ) ]

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

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

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

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

$∴$ $\frac{B P}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

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