【摘 要】
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问题如图1,在△ABC 中,∠BAC= 45°,AD⊥BC于 D,BD=3,DC=2,求AD. 本题虽小,但题目条件简洁明了,且问题内涵丰富,因此细细咀嚼,思路广,趣味浓,值得探究.下面从五个方面予以说
【机 构】
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江苏省泗阳县众兴中学,江苏省泗阳县众兴中学 223700,223700
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问题如图1,在△ABC 中,∠BAC= 45°,AD⊥BC于 D,BD=3,DC=2,求AD. 本题虽小,但题目条件简洁明了,且问题内涵丰富,因此细细咀嚼,思路广,趣味浓,值得探究.下面从五个方面予以说明. 一、巧用面积关系求解 解令AD=x,则由勾股定理得 AB=(x2-9)～(1/(x2-9)),AC=(x2+4)～(1/(x2+4)).作 CE⊥AB于E,则CE=AC·sin45°, 由 S△ABC=1/2BC·AD
The problem is shown in Figure 1. In △ABC, ∠BAC=45°, AD⊥BC in D, BD=3, DC=2, and seek AD. This problem is small, but the conditions are concise and clear, and the problem is rich in content, and therefore fine. Fine chew, a wide range of ideas, interesting, it is worth exploring. The following description from five aspects. First, the clever use of the area relationship to solve the solution AD = x, then by the Pythagorean Theorem AB = (x2-9) ~ (1/ (x2-9)), AC=(x2+4)~(1/(x2+4)). For CE⊥AB at E, then CE=AC·sin45°, by SΔABC=1/2BC·AD
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