AB=AC,AB⊥AC

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AB=AC,AB⊥AC
AB=AC,AB⊥AC

AB=AC,AB⊥AC
AC=AB,外加垂直条件得出∠ABC=∠ACB=45°,∠BAC=90°
梯形上下底平行,∠ABC+BAD=180°计算得∠BAD=135°
设AB=AC=1,根据勾股定理得BC=BD=√2
根据正弦定理
AB/sin(∠ADB)=BD/sin(∠BAD)
1/sin(∠ADB)=√2/sin(135°)=√2/（√2/2）=2
得出sin(∠ADB)=0.5查表可得∠ADB=30°梯形上下底平行∠ADB=∠DBC=30°
∠DBC=30°,BC=BD,∠DCB=∠CDB=75°

AB=AC,AB⊥AC AB=AC, 如图,已知AB⊥BD,AC⊥AB,AB=AC,求证:BD=CD AB⊥BC,AB=AC=2是什么三角形向量AB+向量AC=?,向量AB-向量AC=? 向量ab-ac=向量ab+ca 如图,AB=AC, 等腰三角形ABC,AB=AC,：三角形ABC,AB=AC, 向量填空,三角型abc若|ab+ac|=|ab-ac|则ab乘ac= ab⊥ac.ad⊥ae .ab=ac .ad=ae .说明 be⊥cd DA⊥AB EA ⊥AC AB=AD AC=AE 求证 DC=BE CE⊥AB.DF⊥AB.AC‖BD.AC=DB 求证CE=DF 如图,AB=AC,BD⊥AC,CE⊥AB求证BE=CD 已知,如图,AB=AC,CE⊥AB,BF⊥AC求证OA=OD 已知AB⊥BD,AC⊥CD,AB=AC,求证BD=CD 如图,AB⊥AC,AC⊥DC,AD=AB.试说明AD∥CB. 已知：如图,AB=BD,AC⊥CD,AB=AC.求证：BD=CD