Question

在三角形ABC中,AB=2根号5,AC=4,BC=2,以AB为边做等腰直角三角形,求CD长（三种情况）

Answer 1

(1)当AB为斜边时,BD=根号10,CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC+45°）或

CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC-45°）解得CD=根号2或3根号2

（2）当AB为直角边时,BD=2根号5,CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC+90°）或CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC-90°）解得CD=2根号2或2根号10

（3）当AB为直角边时,BD=2根号10,CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC+45°）或CD^2=BC^2+BD^2-2BC*BD*cos(∠ABC+45°）解得CD=2根号13或2根号10