Question

如图线段AC,BD为四边形ABCD对角线.已知,∠ABC=∠ADC=90°,AD=DC吗,AD=DC,tan∠A

Answer 1

延长BC到E,使CE=AB,连接DE【或将△DAB以点D为轴,逆时针旋转90°,使AD与CD重合...】∵∠ABC=∠ADC=90°∴∠BAD+∠BCD=180°∵∠DCE+∠BCD=180°∴∠BAD=∠DCE又∵AB=CE,AD=DC∴△DAB≌△DCE（SAS）∴∠ADB=∠CDE,BD=DE=6∵∠ADB+∠BDC=∠ADC=90°∴∠BDE=∠CDE+∠BDC=90°则△BDE为等腰直角三角形∴BE=6√2∵tan∠ACB=AB/BC=CE/BC=5/7∴BC=7/12BE=7√2/2,AB=CE=5/12BE=5√2/2∵AC²=AB²+BC²=37∴CD=√(AC²/2)=(√74)/2