Question

在三角形ABC中,角平分线BD与CE交于F,已知∠A=55°求∠EFD的度数

Answer 1

∵∠A=55°

∴∠ABC+∠ACB=180°-∠A=125°

∵BD、CE分别平分∠ABC和∠ACB

∴∠FBC=1/2∠ABC,∠ECB=1/2∠ACB

∴∠FBC+∠ECB=1/2(∠ABC+∠ACB)=1/2×125°=62.5°

∴∠CFB=180°-（∠FBC+∠ECB）=180°-62.5°=117.5°

∴∠EFD=∠CFB=117.5°