如图,AB=AE,AB⊥AE,AD=AC,AD⊥AC,点M为BC的中点.求证:DE=2AM,(中线倍长法)
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如图,AB=AE,AB⊥AE,AD=AC,AD⊥AC,点M为BC的中点.求证:DE=2AM,(中线倍长法)
如图,AB=AE,AB⊥AE,AD=AC,AD⊥AC,点M为BC的中点.求证:DE=2AM,(中线倍长法)
如图,AB=AE,AB⊥AE,AD=AC,AD⊥AC,点M为BC的中点.求证:DE=2AM,(中线倍长法)
延长AM至N,使MN=AM,则ABNC是平行四边形.
∠CAN=∠ANB
由已知得∠EAD+∠BAC=180°
△ABN中,
∠ABN+∠BAN+∠ANB=180°
所以
∠ABN+∠BAN+∠CAN=180°
即
∠ABN+∠BAC=180°
又∠EAD+∠BAC=180°
所以∠ABN=∠EAD
又BN=AC=AD,BA=AE
所以,△BNA≌△ADE
所以,NA=DE
所以,2AM=DE
你怎么知道它是平行四边形?
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