Question

已知,图①中,P是三角形ABC中角B和角C的平分线的交点；图②中,P是三角形ABC中角B的平分线与三角形ABC的外角角A

Answer 1

图①中,∠P＝90°+½∠A．

证明：在△BPC中∠P＝180°-½﹙∠A BC+∠ACB﹚=180º﹣½﹙180°-∠A﹚=90°+∠A.

图②中,∠P＝½∠A．

证明：在△BPC中,

∠P=180°-½﹙∠ABC+ACE﹚-∠ACB

= 180°-½[∠ABC+﹙∠ABC+∠A﹚]-﹙180°-∠ABC-∠A﹚

＝180°-½[2∠ABC+∠A]﹣180°+∠ABC+∠A

＝180°-∠ABC-½∠A﹣180°+∠ABC+∠A

=½∠A．

图③中,∠P＝90°-½∠A

证明：在△BPC中,

∠P= 180°-½﹙∠FBC+∠ECB﹚= 180°-½﹙180°-∠ABC+180°-∠ACB﹚

=180°-½[360°-﹙∠ABC+∠ACB﹚]

=180°-½[360°-﹙180°-∠A﹚]

=180°-½[180°+∠A﹚]

=90-½∠A﹚