Question

In the adjoining figure, $ABCD$ is a quadrilateral with $BA$ parallel to $CD$. $AC$ and $BD$ meet at $X$, where $CX=8cm$ and $XA=10cm$
If $BD=27cm$, then find the length of $BX$

Answer 1

We have,

$\begin{array}{c}\because BA\parallel CD\\ \Delta B\times A\sim \Delta D\times C\\ \Rightarrow \frac{BX}{DX}=\frac{AX}{CX}=\frac{10}{8}=\frac{5}{4}\\ \Rightarrow \frac{BX}{\left(BD-BX\right)}=\frac{5}{4}\\ \Rightarrow \frac{BX}{\left(27-BX\right)}=\frac{5}{4}\\ \Rightarrow 4BX=135-5BX\\ \Rightarrow BX=15cm\end{array}$