Question

等腰梯形ABCD中,AD平行于BC,AC等于BC,AE是高,AE=AD=3,求AC

Answer 1

作DF垂直BC于F.

梯形为等腰梯形,则BE=CF;EF=AD=3.

设BE=CF=X,则BC=2X+3=AC;

AC^2=AE^2+CE^2,即:(2X+3)^2=9+(X+3)^2,解之得:X=1或-3.

X=-3,不合题意,舍去.

所以,AC=2X+3=5.