如题所述
证明:
∵BD⊥AC,
∴∠BDC=∠ADB=90°,
∴BC²=BD²+CD²(勾股定理),
BD²=AB²-AD²,
∵AB=AC,
∴BD²=AC²-AD²,
∴BC²=AC²-AD²+CD²,
∵AC²=(AD+CD)²=AD²+CD²+2AD·CD,
∴BC²=AD²+CD²+2AD·CD-AD²+CD²
=2CD²+2AD·CD
=2CD(CD+AD)
=2CA·CD