作业帮在RT三角形ABC中,∠BAC=90°,AB=AC,分别过BC做过点A的直线的垂线BD CE,.若BD=14cm,CE=3cm,则DE=多少cm
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在RT三角形ABC中,∠BAC=90°,AB=AC,分别过BC做过点A的直线的垂线BD CE,.若BD=14cm,CE=3cm,则DE=多少cm
在RT三角形ABC中,∠BAC=90°,AB=AC,分别过BC做过点A的直线的垂线BD CE,.若BD=14cm,CE=3cm,则DE=多少cm
在RT三角形ABC中,∠BAC=90°,AB=AC,分别过BC做过点A的直线的垂线BD CE,.若BD=14cm,CE=3cm,则DE=多少cm
14+3=17
由已知∠BAC=90,BD、CE垂直DE,
∠DAB+∠BAC+∠CAE=180
∠DAB+∠CAE=180-90=90
因为∠DAB+∠DBA=90
∠CAE+∠ECA=90,
所以
∠DBA=∠CAE,∠DAB=∠ACE
sin∠DBA=DA/AB,sin∠CAE=CE/AC,sin∠DAB=BD/AB,sin∠ACE=AE/AC
所以DA/AB=CE/AC,BD/AB=AE/AC
因为AB=AC,
所以DA=CE=3,BD=AE=14,
DE=DA+AE=3+14=17
题目没错？
分别过BC做过点A的直线的垂线BD CE什么意思啊，谁垂直于谁
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