Question
$A B C D$ is a parallelogram and $E$ is a point on $B C$. If the diagonal $B D$ intersect $A E$ at $F$, prove that $A F \times F B=E F \times F D$.
✓
Answer
Given: $A B C D$ is a parallelogram and $E$ is point on $B C$. Diagonals $D B$ intersects $A E$ at $F$.
To Prove: $AF \times FB = EF \times FD$
Proof: In $\triangle\text{AFD}$ and $\triangle\text{EFD}$
$\angle\text{AFD}=\angle\text{EFB}$ $($vertically opposite $\angle\text{s})$
$\angle\text{DAF}=\angle\text{BEF}$ $($ Alternate $\angle\text{s})$
$\therefore\triangle\text{AFD}\approx\triangle\text{EFD}$ [By AAA similarity]
$\therefore\frac{\text{AF}}{\text{EF}}=\frac{\text{FD}}{\text{FB}}$
$\text{AF}\times\text{FB}=\text{EF}\times\text{FD}$
Hence proved.
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