运动探究| 2 |
| 2 |
如图,在Rt△ABC中,∠ACB=90°,CD平分∠ABC,CE是AB边上的中线,CF⊥AB.
在△ABC中,AC=3,BC=4,AB=5.点D是AB的中点.求CD的长.
如图,△ABC和△ECD都是等腰直角三角形,∠ACB=∠DCE=90°,D为AB边上一点.
如图·ABCD中,∠ADC=78°,AF⊥BC于F,AF交BD于E,若DE=2AB,则∠AED=
In Fig,In the Rt△ABC,∠ACB=90°,∠A=30°,CD is the bisector to∠ACB,MD is the perpendicular to BA and MD through the midpoint of segment AB,then∠CDM=