$ KB = AK$ and $ CJ = JA$ $ \Longrightarrow$ $ \angle FKB = 2 \angle A = \angle CJE.$ In addition, $ \angle BFK = \angle BFC = \angle C - \angle CBF = \angle C - \angle CBP = \angle C - \angle CAP =$ $ = (\angle C - \angle A) + (\angle A - \angle CAP) = \angle BCJ + \angle PAB = \angle BCJ + \angle PCB =$ $ = \angle PCJ = \angle ECJ.$ Triangles $ \triangle BFK \sim \triangle ECJ$ are therefore similar, having equal angles $ \Longrightarrow$ $ \frac{EC}{BF} = \frac{JE}{KB} = \frac{CJ}{FK},\ \ \left(\frac{EC}{BF}\right)^2 = \frac{JE}{KB} \cdot \frac{CJ}{FK} = \frac{JE \cdot JA}{AK \cdot FK} = \frac{AJ \cdot JE}{AK \cdot KF}.$