Question

已知:如图,三角形ABC,∠ACB=90°,CD垂直于AB于点D,AD等于3BD,求证:∠A=30°

Answer 1

∠ACB=90°,CD垂直于AB于点D

所以∠BCD+∠DCA=90

∠DCA+∠DAC=90

∴∠BCD=∠DAC

∴△BCD∽△CAD

BD/CD=CD/AD

设BD=t

则AD=3t

CD^2=BD*AD=3t^2

CD=根号3t

AD/CD=3t/根号3t=根号3

所以在直角△ACD中∠A=30°