Question

如图,在RT三角形ABC中,角C=90度,AC=8,BC=6,按图中所示方法将三角形BCD沿BD折叠,使点C落在边AB上

Answer 1

AB=√(AC^2+BC^2)=10;BC'=BC=6,则AC'=4.

∠BC'D=∠C=90°,则∠AC'D=∠C=90°;

又∠A=∠A,故⊿AC'D∽⊿ACB,AC'/AC=AD/AB.

即:4/8=AD/10,AD=5,CD=AC-AD=3,BD=√(CD^2+BC^2)=3√5.