...=AC,∠B=30度,AD⊥Ac,垂足为点A,点D在BC上.求证:CD=2BD
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热心网友
如图所示
∵AB = AC且∠B = 30°
∴∠C = 30°
∴Rt△ADC中AD = 1/2 CD
又∵∠BAD = ∠BAC - 90° = (180° - 30° - 30°) - 90° = 30°
∴△DAB为等腰三角形且 AD = BD
∴CD = 2AD = 2BD
热心网友
解：∵AB=AC
∴∠B=∠C=30°
即∠BAC=120°
又∵AC⊥AD
∴∠CAD=90°
∴∠BAD=∠BAC-∠CAD=30°
∵∠B=∠BAD=30°
∴AD=BD
在Rt△CAD中，∠C=30°
∴AD=2CD 即CD=2AC
∴CD=2BD
∴
热心网友
如图所示
∵AB = AC且∠B = 30°
∴∠C = 30°
∴Rt△ADC中AD = 1/2 CD
又∵∠BAD = ∠BAC - 90° = (180° - 30° - 30°) - 90° = 30°
∴△DAB为等腰三角形且 AD = BD
∴CD = 2AD = 2BD
