Benchmarks

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总结帝笔记—初三寒假班第三讲

50

$$\backslash because\; DE\; \backslash parallel\; BC\; \backslash quad\; \backslash frac\{AN\}\{AM\}\; =\; \backslash frac\{AD\}\{AB\}\; =\; \backslash frac\{DN\}\{BM\}$$

$$\backslash triangle\; ADN\; \backslash sim\; \backslash triangle\; ABM\; \backslash quad\; \backslash triangle\; ANE\; \backslash sim\; \backslash triangle\; AMC\; \backslash quad\; \backslash frac\{AN\}\{AM\}\; =\; \backslash frac\{AE\}\{AC\}\; =\; \backslash frac\{NE\}\{MC\}$$

$$\backslash because\; AN\; =\; 2NM$$

$$\backslash therefore\; \backslash frac\{AN\}\{AM\}\; =\; \backslash frac\{2\}\{3\}\; \backslash quad\; \backslash text\{故两组相似的相似比\}$$

均为 $\backslash frac\{2\}\{3\}$

$$\backslash therefore\; \backslash frac\{S\_\{\backslash triangle\; ADN\}\}\{S\_\{\backslash triangle\; ABM\}\}\; =\; \backslash left(\backslash frac\{2\}\{3\}\backslash right)^2\; =\; \backslash frac\{4\}\{9\};\; \backslash quad\; \backslash frac\{S\_\{\backslash triangle\; ANE\}\}\{S\_\{\backslash triangle\; AMC\}\}\; =\; \backslash left(\backslash frac\{2\}\{3\}\backslash right)^2\; =\; \backslash frac\{4\}\{9\}$$

$$\backslash text\{设\; \}\; S\_\{\backslash triangle\; ADN\}\; =\; 4k\; \backslash quad\; S\_\{\backslash triangle\; ANE\}\; =\; 4t$$

$$\backslash therefore\; S\_\{\backslash triangle\; ABM\}\; =\; 9k\; \backslash quad\; S\_\{\backslash triangle\; AMC\}\; =\; 9t$$

$$\backslash therefore\; S\_\{\backslash triangle\; ABC\}\; =\; S\_\{\backslash triangle\; ABM\}\; +\; S\_\{\backslash triangle\; AMC\}\; =\; 9k\; +\; 9t$$

$$\backslash because\; S\_\{\backslash triangle\; DMN\}\; =\; \backslash frac\{1\}\{2\}\; S\_\{\backslash triangle\; ADN\}\; \backslash quad\; (\backslash text\{等高不等底\})$$

$$S\_\{\backslash triangle\; MNE\}\; =\; \backslash frac\{1\}\{2\}\; S\_\{\backslash triangle\; ANE\}\; \backslash quad\; \backslash text\{同理\}$$

$$\backslash therefore\; S\_\{\backslash triangle\; DMN\}\; =\; 2k;\; \backslash quad\; S\_\{\backslash triangle\; MNE\}\; =\; 2t$$

$$\backslash therefore\; S\_\{\backslash triangle\; DME\}\; =\; S\_\{\backslash triangle\; DMN\}\; +\; S\_\{\backslash triangle\; MNE\}\; =\; 2k\; +\; 2t$$

$$\backslash therefore\; \backslash frac\{S\_\{\backslash triangle\; DME\}\}\{S\_\{\backslash triangle\; ABC\}\}\; =\; \backslash frac\{2k+2t\}\{9k+9t\}\; =\; \backslash frac\{2(k+t)\}\{9(k+t)\}\; =\; \backslash frac\{2\}\{9\}$$

口嗨不?

6. 在四边形ABCD中, $AD\; \backslash parallel\; BC$ , AB与CD不平行
根据图中数据, 若BA、CD延长后交于点M, 则
$\backslash triangle\; MBC$ 的周长为____.

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