Question

ABCD is a parallelogram. E is a point on BC and the diagonal BD intersects AE at F. Prove that DF x FE = FB x FA.

Answer 1

Given that, the diagonal $BO$ of parallelogram $ABCD$ intersects the segment $AE$ at $F,$ where $E$ is any point on $BC$.

To prove: $DF\times EF=FB\times FA$

Based on the given information, we can draw the figure shown above.

Now, in $△AFD$ and $△BFE$,

$\Rightarrow$ $\angle FAD=\angle FEB$ [ Alternate angles ]

$\Rightarrow$ $\angle AFD=\angle BFE$ [ Vertically opposite angles ]

$\therefore$ $△ADF\sim △BFE$ [ By $AA$ similarity ]

$\therefore$ $\frac{DF}{FA}=\frac{FB}{EF}$

$\therefore DF\times EF=FB\times FA\left[Henceproved\right]$