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在凸四边形ABCD中,角BCD=角ADC=90度,点E在边CD上,且角AEB=90度,AE=BE,点M是边AB的中点,MC交BE于点P,MD交AE于点Q,ME交PQ于点R,求证MR大于等于RE,并确定等号何时成立.
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证：在平面上取一点P,使角DPE=角DBE,角BPC=角BAC.则A、B、C、P共圆,B、E、D、P共圆.角EDP+角EBP=180度,角ACP=角ABP 则角FDP+角FCP=180度,C、F、D、P共圆.所以ΔABC,ΔCDF,ΔBDE的外接圆三圆共点
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In prominent figures primers, Kok BCD= Kok ADC=90 degrees, the point e in the margin CD, and Kok AEB=90 degrees AE=BE, point M is at the mid-point of AB, MC for be in point P,MD to amend in point Q,ME to PQ in point R, see displays more than equal to re, and to determine when to set up sign.
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Session 04 – Stem cells | Human Reproduction | Oxford Academic
Issue suppl_1
Session 04 – Stem cells
Human Reproduction, Volume 20, Issue suppl_1, 1 June 2005, Pages i5–i7,
Session 04 – Stem cells, Human Reproduction, Volume 20, Issue suppl_1, 1 June 2005, Pages i5–i7,
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