Question

A straight line cuts the $x$-axis at point $A(1,0)$ and $y$-axis at point $B$, such that $\angle OAB=\alpha ,(\alpha >\pi /4)$.
$C$ is the middle point of $AB$. If $B^{\prime }$ is the mirror image of $B$ with respect to line $OC$ and $C^{\prime }$ is a mirror image of point $C$ with respect to line $B{B}^{\prime }$, then the ratio of the areas of triangle $AB{B}^{\prime }$ and $B{B}^{\prime }{C}^{\prime }$ is

Answer 1

$\because D$ is the mid-point of $B{B}^{\prime }$ and $C$ is the mid-point of $AB$
So line $CD\parallel A{B}^{\prime }$ and $CD=\frac{1}{2}A{B}^{\prime }$
Now area of $△AB{B}^{\prime }=\frac{1}{2}A{B}^{\prime }\times B{B}^{\prime }$
$=A{B}^{\prime }\times BD$
Area of $△B{C}^{\prime }{B}^{\prime }=\frac{1}{2}B{B}^{\prime }\times D{C}^{\prime }$
$=\frac{1}{2}\times 2BD\times CD(\because CD=C{D}^{\prime })$
$=\frac{1}{2}BD\times A{B}^{\prime }$
Hence, required ratio is $2$