As in right figure,in a quadrilateral ABCD,we have its diagonal AC bisects∠DAB,and AB=21,AD=9,BC=DC=10,then the distance from point C to line AB is
解:过C作CE⊥AD,CF⊥AB,| AF2+CF2 |
| 152+82 |
(2013·咸宁)如图,在平面直角坐标系中,以O为圆心,适当长为半径画弧,交x轴于点M,交y轴于点N,再分别以点M、N为圆心,大于| 1 |
| 2 |
如图,在△ABC中,∠C=90°,∠BAC的平分线AD交BC于D,若BD=15,BD:CD=5:3,AB=30,则△ABD的面积是
如图,在△ABC中,∠C=90°,BC=12cm,∠A的平分线交BC于D,DB=8cm,则点D到斜边AB的距离为