Question

如图点O是三角形ABC三条角的平分线的交点,过点O作OG垂直BC垂足为点G,求证角1=角2.

Answer 1

∵ ∠COG=90°-½∠BCA

又∵∠DOB=∠AOE=180°-∠OAE-∠OEA

=180°-∠OAE-(∠OBC+∠BCE)

=180°-∠OAE-∠OBC-∠BCE

=180°-½∠BAC-½∠ABC-∠BCE

=180°-½（180°-∠BCA）-∠BCE

=180°-90°-½∠BCA

=90°-½∠BCA

∴∠DOB=∠COG

（∠DOB,∠COG就是你的角一角二）