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

Given:

 In $\triangle ABC$, $PQ$ is a line segment intersecting $AB$ at $P$ and $AC$ at $Q$ such that $PQ\parallel BC$ and $PQ$ divides $\triangle ABC$ into two parts equal in area.

To do:

We have to find $\frac{BP}{AB}$.

Solution:

$PQ\parallel BC$ and $ar(\triangle APQ)=ar(trapezium\ BCQP)=\frac{1}{2}ar(\triangle ABC)$

In $\triangle APQ$ and $\triangle ABC$,

$\angle PAQ=\angle BAC$   (Common angle)

$\angle APQ=\angle ABQ$   (Corresponding angles)

Therefore,

$\triangle APQ \sim\ \triangle ABC$  (By AA similarity)

This implies,

$\frac{ar(\triangle APQ)}{ar(\triangle ABC)}=\frac{(AP)^2}{(AB)^2}$   (Area of similar triangles theorem)

$\frac{\frac{1}{2}ar(\triangle ABC)}{ar(\triangle APQ)}=(\frac{AP}{AB})^2$ 

$(\frac{AP}{AB})^2=\frac{1}{2}$

$\frac{AP}{AB}=\frac{1}{\sqrt{2}}$

$AP=\frac{AB}{\sqrt2}$

$BP=AB-AP$

$BP=AB-\frac{AB}{\sqrt2}$

$BP=AB(\frac{\sqrt2-1}{\sqrt2})$

Therefore,

$\frac{BP}{AB}=\frac{AB(\frac{\sqrt2-1}{\sqrt2})}{AB}$

$=\frac{AB(1-\sqrt2)}{AB}$

$=\frac{\sqrt2-1}{\sqrt2}$

The value of $\frac{BP}{AB}$ is $\frac{\sqrt2-1}{\sqrt2}$.

Tutorialspoint

Updated on: 10-Oct-2022

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