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1、做AM⊥BC
∵AB=AC
∴BM=CM=1/2BC
∴AM²=AD²-DM²
AM²=AB²-BM²
∴AD²-DM²=AB²-BM²
AD²-AB²=DM²-BM²=(DM+BM)(DM-BM)=(DM+CM)(DM-BM)=CD×BD
即AD²-AB²=BD·CD
2、做AM⊥BC(D在BM上)
∵AB=AC
∴BM=CM=1/2BC
∴AM²=AD²-DM²
AM²=AB²-BM²
∴AD²-DM²=AB²-BM²
AB²-AD²=BM²-DM²=(BM+DM)(BM-DM)=(CM+DM)(BM-DM)=CD×BD
即AB²-AD²=BD·CD
∵AB=AC
∴BM=CM=1/2BC
∴AM²=AD²-DM²
AM²=AB²-BM²
∴AD²-DM²=AB²-BM²
AD²-AB²=DM²-BM²=(DM+BM)(DM-BM)=(DM+CM)(DM-BM)=CD×BD
即AD²-AB²=BD·CD
2、做AM⊥BC(D在BM上)
∵AB=AC
∴BM=CM=1/2BC
∴AM²=AD²-DM²
AM²=AB²-BM²
∴AD²-DM²=AB²-BM²
AB²-AD²=BM²-DM²=(BM+DM)(BM-DM)=(CM+DM)(BM-DM)=CD×BD
即AB²-AD²=BD·CD
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