Question

27.(9分)-|||-(1)如图,在+ABC+中,+ACB=90+,CDAB,+垂足为D.求证+BC^2=BDBA.-||

Answer 1

问：27.(9分)-|||-(1)如图,在+ABC+中,+ACB=90+,CDAB,+垂足为D.求证+BC^2=BDBA.-||

答：因为 $\angle ACB=90^\circ$，所以根据勾股定理，$AC^2=AB^2+BC^2$。$\because CD\perp AB$，$\therefore \angle ADB=90^\circ$。同样根据勾股定理，$AD^2=AB^2+BD^2$。将两个式子相减，可得：$AC^2-AD^2=BC^2-BD^2$即 $BC^2=BD^2+(AC^2-AD^2)$又因为 $AC=AD$，所以 $AC^2=AD^2$，代入上式可得：$BC^2=BD^2$因此，$BC^2=BDBA$。