Question

如题24图,⊙O是Rt△ABC的外接圆,∠ABC=90°,弦BD=BA,AB=12,BC=5,

BE⊥DC交DC的延长线于点E.

(1)求证：∠BCA=∠BAD;

(2)求DE的长;

(3)求证:BE是⊙O的切线.

Answer 1

(1)∵AB=DB,∴∠BDA=∠BAD,又∵∠BDA=∠BCA,∴∠BCA=∠BAD.

(2)在Rt△ABC中，AC=,易证△ACB∽△DBE,得,

∴DE=

(3)连结OB,则OB=OC,∴∠OBC=∠OCB,

∵四边形ABCD内接于⊙O,∴∠BAC+∠BCD=180°,

又∵∠BCE+∠BCD=180°,∴∠BCE=∠BAC,由(1)知∠BCA=∠BAD,∴∠BCE=∠OBC,∴OB∥DE

∵BE⊥DE,∴OB⊥BE,∴BE是⊙O的切线.