如图,在三角形ABC中,AB=AC,角BAC=90度,点D在BC上,试说明BD^2+CD^2=2AD^2

过A点作AE^BC于E
则在RtΔ ADE中,AD2=DE2+AE2
又∵ΔABC为等腰RtΔ
∴ AE=BE=CE
又 BD2+CD2=(BE－DE)2+(CE+DE)2
=BE2+CE2+2DE2
=2AE2+2DE2
=2AD2
即 BD2+CD2=2AD2