解:取 CD 中点E，连接
AE:<BAC=0.2，E为 CD 中点∴AE=BE=CE∴LEAC=∠C∴ㄥAEB= ∠EAC+∠C=2/C:AB=DC/2·AB= AE∴∠B= ∠AEB∴∠B=2∠C∴<BAE=180-ZB-
LAEB = 180-4∠C∴∠BAC =
ZBAE+ZEAC = 180-4∠C+ㄥC =180-3∠C:∠C=20∴∠KFC= V50