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总结帝笔记—初三寒假班第三讲

55.

\therefore S_{\triangle COD} = 4 \quad \therefore S_{\triangle AOD} = 2 \quad 9 \text{故 } S_{\text{梯形}ABCD} = 1+2+2+4 = 9

8. 如图 \square ABCD , F 为 BC 中点, 延长 AD 至 E , 使 DE:AD=1:3 , 连 EF 交 DC 于点 G , 则 S_{\triangle DEG}:S_{\triangle CFG}=

Diagram showing a parallelogram ABCD. E is an external point such that DE is an extension of AD. F is the midpoint of BC. Line segment EF intersects DC at G.
Diagram showing a parallelogram ABCD. E is an external point such that DE is an extension of AD. F is the midpoint of BC. Line segment EF intersects DC at G.

解:

\therefore \text{四边形 } ABCD \text{ 为平行四边形} \therefore DE \parallel FC, AD = BC \therefore DE:AD=1:3 \quad \therefore DE:BC=1:3 \text{又} \because F \text{ 为 } BC \text{ 中点} \quad \therefore DE:FC=2:3 \text{又} \therefore DE \parallel FC \quad \therefore \triangle DEG \sim \triangle CFG \therefore \frac{DE}{FC} = \frac{2}{3} \text{ 为相似比} \therefore \frac{S_{\triangle DEG}}{S_{\triangle CFG}} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}

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