What is the value of $a+2b$ if $\frac{OP}{EF}=\frac{a}{}b$ in $ABCD$ rectangle?

Question

BD is diagonal in ABCD rectangle and E, F is the mid point of BC, CD respectively. Line BD intersects AE, AF at O, P respectively. If $\dfrac{OP}{EF} = \dfrac{a}b$, then a+2b=?

Answer 1

I'll assume that $a$ and $b$ are the width and height of the rectangle.

Extend $AD$, $EF$ and $AB$ until they meet in points $G,H$. You can easily prove that triangles $BEG$, $CEF$ and $DHF$ are congruent. It follows that $GE=EF=FH$. Lines $EF$ and $BD$ are parallel so by Thales $BO=OP=PD$ too. If the length of diagonal BD is $d$ it means that $OP=d/3$.

On the other side, $EF=d/2$ and from the text of the prboelm we know that:

$$\frac{OP}{EF}=\frac{d \over 3}{d \over 2}=\frac23=\frac ab \implies 3a=2b$$

It follows that $a+2b=4a$.