证明:
作AE⊥BC 于E;则∠AEC=90°
∵ AB =AC
∴ AE = EC =1/2BC
∵ BD⊥AC
∴ ∠BDC = ∠AEC = 90°;∠C=∠C
∴ ΔBDC ∽ΔAEC ==> AC/BC = EC/CD
==> BC*EC = CA*CD
==> BC *1/2*BC =CA*CD
∴ BC²=2CA•CD
结论成立
证明:
作AE⊥BC 于E;则∠AEC=90°
∵ AB =AC
∴ AE = EC =1/2BC
∵ BD⊥AC
∴ ∠BDC = ∠AEC = 90°;∠C=∠C
∴ ΔBDC ∽ΔAEC ==> AC/BC = EC/CD
==> BC*EC = CA*CD
==> BC *1/2*BC =CA*CD
∴ BC²=2CA•CD
结论成立