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证明:
∠CGH
=90°-1/2∠ACB
=90°-1/2[180°-(∠ABC+∠CAB)]
=1/2(∠ABC+∠CAB)
=1/2∠ABC+1/2∠CAB
=∠ABG+∠BAG
又∵∠ABG+∠BAG=∠BGD
∴∠CGH=∠BGD
得证!
∠CGH
=90°-1/2∠ACB
=90°-1/2[180°-(∠ABC+∠CAB)]
=1/2(∠ABC+∠CAB)
=1/2∠ABC+1/2∠CAB
=∠ABG+∠BAG
又∵∠ABG+∠BAG=∠BGD
∴∠CGH=∠BGD
得证!
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