证明:连FA,有FA=FD,证△FBA∽△FAC
∠FDA=∠FAD=∠FAB+∠BAD ∠BAD=∠CAD
∠FDA=∠FCA+∠CAD= ∠FAB +∠BAD,则∠FCA=∠FAB
另∠AFB=∠CFA 所以△FBA∽△FAC(AAA)
FA:FC=FB:FA 即FD的平方=FB*FC.
证明:连FA,有FA=FD,证△FBA∽△FAC
∠FDA=∠FAD=∠FAB+∠BAD ∠BAD=∠CAD
∠FDA=∠FCA+∠CAD= ∠FAB +∠BAD,则∠FCA=∠FAB
另∠AFB=∠CFA 所以△FBA∽△FAC(AAA)
FA:FC=FB:FA 即FD的平方=FB*FC.