设BB1与AC交点是E (BB1⊥AC, BE=B1E)
AC=√﹙AB²+BC²﹚=3㎝
EB1=2√2×1/3=2√2/3㎝
∴AE=√﹙AB1²﹣BB1²﹚=8/3
CE=AC-AE=1/3㎝
∴DB1=AE﹣CE=7/3㎝