如图:已知在三角形ABC中,角BAC=90度,AB=AC,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE如图:已知在三角形ABC中,角BAC=90度,AB=AC,点D在边BC上,以AD为边作正方形ADEF,联结CF,C
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![如图:已知在三角形ABC中,角BAC=90度,AB=AC,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE如图:已知在三角形ABC中,角BAC=90度,AB=AC,点D在边BC上,以AD为边作正方形ADEF,联结CF,C](/uploads/image/z/12670252-52-2.jpg?t=%E5%A6%82%E5%9B%BE%3A%E5%B7%B2%E7%9F%A5%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92BAC%3D90%E5%BA%A6%2CAB%3DAC%2C%E4%BB%A5AD%E4%B8%BA%E8%BE%B9%E4%BD%9C%E6%AD%A3%E6%96%B9%E5%BD%A2ADEF%2C%E8%81%94%E7%BB%93CF%2CCE%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81FC%E5%9E%82%E7%9B%B4BC%EF%BC%882%EF%BC%89%E8%8B%A5BD%3DAC%2C%E6%B1%82%E8%AF%81%EF%BC%9ACD%3DCE%E5%A6%82%E5%9B%BE%3A%E5%B7%B2%E7%9F%A5%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92BAC%3D90%E5%BA%A6%2CAB%3DAC%2C%E7%82%B9D%E5%9C%A8%E8%BE%B9BC%E4%B8%8A%EF%BC%8C%E4%BB%A5AD%E4%B8%BA%E8%BE%B9%E4%BD%9C%E6%AD%A3%E6%96%B9%E5%BD%A2ADEF%2C%E8%81%94%E7%BB%93CF%2CC)
如图:已知在三角形ABC中,角BAC=90度,AB=AC,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE如图:已知在三角形ABC中,角BAC=90度,AB=AC,点D在边BC上,以AD为边作正方形ADEF,联结CF,C
如图:已知在三角形ABC中,角BAC=90度,AB=AC,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE
如图:已知在三角形ABC中,角BAC=90度,AB=AC,点D在边BC上,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE
如图:已知在三角形ABC中,角BAC=90度,AB=AC,以AD为边作正方形ADEF,联结CF,CE(1)求证FC垂直BC(2)若BD=AC,求证:CD=CE如图:已知在三角形ABC中,角BAC=90度,AB=AC,点D在边BC上,以AD为边作正方形ADEF,联结CF,C
还需要补充说明:D、E在AC的两侧,否则需要求证的结论不成立.
[证明]
(1)
∵AB=AC、∠BAC=90°,∴∠ACD=45°.
∵ADEF是正方形,∴∠AED=45°,又∠ACD=45°,∴A、D、C、E共圆.
∵ABCD是正方形,∴A、B、C、D共圆,又A、D、C、E共圆,∴A、D、C、F共圆.
∵ABCD是正方形,∴AD⊥AF,而A、D、C、F共圆,∴FC⊥BC.
(2)
∵AB=AC、BD=AC,∴AB=BD,∴∠BAD=∠BDA.
∵ABCD是正方形,∴∠ADE=90°,∴∠BDA+∠CDE=90°,又∠BAD+∠CAD=90°,
∴∠BAD+∠CAD=∠BDA+∠CDE,而∠BAD=∠BDA,∴∠CAD=∠CDE.
∵A、D、C、E共圆,∴∠CAD=∠CED,又∠CAD=∠CDE,∴∠CED=∠CDE,∴CD=CE.
有图没?D在哪里