Rt三角形ABC中,角C=90度,AC=BC,AD是角A的平分线,求AC+CD=AB

设AC＝BC＝a,则
AB＝sqrt(AC^2＋BC^2)＝sqrt(2)a
∵AD是∠A的平分线
∴BD/CD＝AB/AC＝sqrt(2)
又BD＋CD＝BC＝a
∴CD＝1/[sqrt(2)＋1]·a＝a/[sqrt(2)＋1]
∴AC＋CD＝a＋a/[sqrt(2)＋1]＝a[sqrt(2)＋1＋1]/[sqrt(2)＋1]＝sqrt(2)a＝AB