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乌贼
posted 2022-5-4 23:49
如图:
$ BM\px CF,MN\px CE $\[
\ S_{\triangle MFN}=\dfrac{1}{2}S_{BAFL}=S_{AFNG}\riff S_{\triangle MAK}=S_{\triangle NGK} \]又平行四边形$ KNPG $中\[S_{\triangle MAK}= S_{\triangle NKG}=\dfrac{1}{2}S_{KNPG}=S_{KHEG} \]有\[ S_{\triangle CDF}+S_{\triangle BED}=S_{\triangle BAM}+S_{\triangle BED}=S_{\triangle BHM}=S_{\triangle CEF}=12 \] |
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