Question

ABCD is parallelogram and E is a point on BC.  If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.  

  

Answer 1

We have: 

∠𝐴𝐹𝐷= ∠𝐸𝐹𝐵 (𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠)
∵ DA || BC
∴ ∠𝐷𝐴𝐹= ∠𝐵𝐸𝐹 (𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠)
Δ DAF ~ Δ BEF (AA similarity theorem)
⟹ 𝐴𝐹𝐸𝐹=𝐹𝐷𝐹𝐵
Or, AF × FB = FD × EF
This completes the proof