Benchmarks

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总结帝笔记—初三寒假班第三讲

48

本题: $DE\; ext\{\; \}\; ext\{||\}\; ext\{\; \}\; BC$

$$\backslash therefore\; \backslash angle\; 1\; =\; \backslash angle\; 2\; \backslash text\{\; 又\; \}\; \backslash angle\; A\; =\; \backslash angle\; A$$

$$\backslash therefore\; \backslash triangle\; ADE\; \backslash sim\; \backslash triangle\; ABC$$

$$\backslash therefore\; \backslash frac\{AD\}\{AB\}\; =\; \backslash frac\{2\}\{3\}\; \backslash text\{\; 为相似比\}$$

$$\backslash therefore\; \backslash frac\{S\_\{\backslash triangle\; ADE\}\}\{S\_\{\backslash triangle\; ABC\}\}\; =\; \backslash left(\backslash frac\{2\}\{3\}\backslash right)^2\; =\; \backslash frac\{4\}\{9\}$$

$$\backslash text\{设\; \}\; S\_\{\backslash triangle\; ADE\}\; =\; 4X,\; S\_\{\backslash triangle\; ABC\}\; =\; 9X$$

$$\backslash therefore\; S\_\{\backslash text\{四边形\}DBCE\}\; =\; 9X\; -\; 4X\; =\; 5X$$

$$\backslash therefore\; \backslash frac\{S\_\{\backslash triangle\; ADE\}\}\{S\_\{\backslash text\{四边形\}DBCE\}\}\; =\; \backslash frac\{4X\}\{5X\}\; =\; \backslash frac\{4\}\{5\}$$

5. 如图, 在 $\backslash triangle\; ABC$ 中, 点 $D,\; E$ 分别在 $AB$ 和 $AC$ 上,
$DE\; \backslash parallel\; BC$ , $M$ 为 $BC$ 边上一点(不与点 $B,\; C$ 重合). 连 $AM$
交 $DE$ 于点 $N$ . 则

$$\backslash text\{A.\; \}\; \backslash frac\{AD\}\{AN\}\; =\; \backslash frac\{AN\}\{AE\};\; \backslash text\{B.\; \}\; \backslash frac\{BD\}\{MN\}\; =\; \backslash frac\{MN\}\{CE\};\; \backslash text\{C.\; \}\; \backslash frac\{DN\}\{BM\}\; =\; \backslash frac\{NE\}\{ME\};\; \backslash text\{D.\; \}\; \backslash frac\{DN\}\{MC\}\; =\; \backslash frac\{NE\}\{BN\}$$

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