如图,在Rt△ABC中,∠ACB=90°,AC=5,CB=12,AD是△ABC的角平分线,过A、C、D三点的圆与斜边AB交于点E,连接DE。 (1)求证:AC=AE; (2)求△ACD外接圆的半径。 答案:(1)证明:∵∠ACB=90°, ∴AD为直径。 又∵AD是△ABC的角平分线, ∴,∴ ∴AC=AE (2)解:∵AC=5,CB=12, ∴AB= ∵AE=AC=5,∴BE=AB-AE=13-5=8 ∵AD是直径,∴∠AED=∠ACB=90° ∵∠B=∠B,∴△ABC∽△DBE ∴,∴ DE= ∴AD= ∴△ACD外接圆的半径为 …………………(8分14.Many young people took part in _______ trees on Tree Planting Day.( )A.plantingB.plantsC.to plantD.plant