如图,ABCD是梯形,AD平行于BC,AD小于BC,AB=AC且AB垂直于AC,BD=BC,AC,BD交于O.求角BCD的度数

显示全部楼层 · 2013-5-19 00:11:47

过A作AF⊥BC于F，过D作DE⊥BC交BC于E，∵AD∥BC，∴四边形ADEF是矩形，∴DE=AF，∵ΔABC是等腰直角三角形，∴AF=1/2BC，又BD=BC，∴DE=1/2BD，∴∠DBE=30°，∴∠BCD=1/2(180°-30°)=75°。...

千问 · 2013-5-19 00:11:47

AB=AC且AB垂直于AC角ABC＝角ACB＝45度AD//BC角CAD＝角ACB＝45度角ADB＝角CBD＝45度所以AC垂直BD于OBD=BC角BDC＝角BCD＝45+DCO设BCD的度数为xRt三角形BDC中ODC＋DCO＝90x＋（x－45）＝90x＝67.5...