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乌贼
posted 2023-11-12 01:33
Last edited by 乌贼 2023-11-20 00:41如图：
直接算出来就行了，令$ DE=b,AD=b $\[ AG^2=AF^2=AD^2+DF^2+2AD\cdot DF\cdot \cos \angle ADF=a^2+b^2+ab \]\[ DG^2=DB^2=BH^2+DH^2=AB^2-AH^2+DH^2=a^2-(\dfrac{a-b}{2})^2+(\dfrac{a+b}{2})^2=a^2+ab \]则有\[ AG^2=a^2+b^2+ab=DG^2+AD^2 \]即有\[ DG\perp AE \] |
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