Only if part: I think this can be done by proving: $K$ is the midpoint of $AX \implies \frac{AD}{DC} = \sqrt{2}$ Note that by POP, $BL^2 = KL \cdot AL$. Let $KL = a$; then $BL = a\sqrt{2}$. Also, by SAS $\triangle BLK \sim \triangle ALB$. Thus $\angle KBL = \angle BAL$. Let $\angle BAL = \alpha$; then \[\angle BAD = \angle CAD = \angle CBK = \angle BCK = \angle KBL = \alpha\] It follows that $BK$ is the angle bisector of $\angle DBL$, so by Angle Bisector Theorem \[\sqrt{2} = \frac{BL}{KL} = \frac{BD}{DK}\] Since $\triangle BDK \sim \triangle ADC$, \[\frac{BD}{DK} = \frac{AD}{DC} = \sqrt{2}\] If part: This can be done by proving: $\frac{AD}{DC} = \sqrt{2} \implies K$ is the midpoint of $AX$. $BL$ is a tangent, and $\angle KBL$ 'slices off' $\overarc{BK}$; since $\angle BAK$ is an inscribed angle that subtends $\overarc{BK}$ (thanks http://www.mathillustrations.com/figureGallery/index.php?mode=entry&id=139 for helping with terminology), $\angle KBL = \angle BAK$. So $\triangle BLK \sim \triangle ALB$. Let $\angle BAD = \alpha$; then \[\angle BAD = \angle CAD = \angle CBK = \angle BCK = \angle KBL = \alpha\] And once again $BK$ is the angle bisector of $\angle DBL$. Then we can use techniques in the only if part to show $\triangle ADC \sim \triangle BDK$, and $\frac{AD}{DC} = \frac{BD}{DK} = \sqrt{2}$, and $\frac{BL}{KL} = \sqrt{2}$. By POP we have $BL^2 = KL \cdot AL$, so $AL = 2KL$ and the result follows.