Question

如图,三角形ABC中,角ACB=90度,点D在AB上,且角1=角A,求证：角ACD=2角B

Answer 1

∵∠1=∠B+∠BCD,∠1=∠A∴∠A=∠B+∠BCD,又∠ACB=90°∴∠A+∠B=∠ACD+∠BCD=90°∴∠B+∠BCD+∠B=∠ACD+∠BCD∴∠ACD=2*∠B