The parallel line passing through A is BF, which extends from FD to G. It is easy to know that △AEG is similar to △CEF. And because E is the midpoint of AC, we can know that CF=AG from the similar triangles property. It is easy to know that △ADG and △BDF are similar, so AD/BD=AG/BF, so AD*BF=AG*BD, and CF=AG, so AD * BF = BD * CF