Question

$P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $△ABC$ such that $PQ\parallel BC$ and divides the triangle into two parts of equal area. Find $PB:AB$.

• A. $1:\sqrt{2}$
• B. $\left(\sqrt{2}\right):(\sqrt{2}-1)$
• C. $(\sqrt{2}-1):\sqrt{2}$
• D. $1:(\sqrt{2}-1)$

Answer 1

Answer: (C)

$(\sqrt{2}-1):\sqrt{2}$

It is given that $PQ$ divides $△ABC$ into two parts of equal area. So,

$\begin{array}{cc}ar\left(APQ\right)& =ar\left(PBCQ\right)\\ ar\left(APQ\right)+ar\left(APQ\right)& =ar\left(PBCQ\right)+ar\left(APQ\right)\\ 2ar\left(APQ\right)& =ar\left(ABC\right)\\ \frac{ar\left(APQ\right)}{ar\left(ABC\right)}& =\frac{1}{2}\dots \left(i\right)\end{array}$

Now, in $△APQ$ and $△ABC$, we have
$\angle PAQ=\angle BAC$ [Common]
$\angle APQ=\angle ABC$ [Corresponding angles]

So, by $AAA$ criteria of similarity,

$△APQ\sim △ABC$

We know that the areas of similar triangles are proportional to the squares of their corresponding sides.
$\therefore$ $\frac{ar\left(△APQ\right)}{ar\left(△ABC\right)}=\frac{{AP}^2}{{AB}^2}$

$\Rightarrow$ $\frac{A{P}^2}{A{B}^2}=\frac{1}{2}$ [By using $\left(i\right)$]

$\Rightarrow AB=\sqrt{2}\cdot AP$

$\Rightarrow$ $AB=\sqrt{2}(AB-PB)$

$\Rightarrow$ $\sqrt{2}PB=(\sqrt{2}-1)AB$

$\Rightarrow$ $\frac{PB}{AB}=\frac{(\sqrt{2}-1)}{\sqrt{2}}$

$\therefore$ $PB:AB=(\sqrt{2}-1):\sqrt{2}$