1.求证:三角形两个外角平分线所夹的锐角等于90°与第三个内角一半的差.

1.设△ABC中,
CD是∠BAC外角平分线,
AD是∠CAB外角平分线,
CD,AD交于D,
由条件知：∠ACD=1/2（∠A+∠B）,
∠CAD=1/2（∠B+∠C）,
∴∠CDA=180°-1/2（∠A+∠B）-1/2（∠B+∠C）
=180°-1/2(∠A+∠B+∠C)-1/2∠B
=180°-90°-1/2∠B
=90°-1/2∠B.
2.设△ABC中,BD是∠ABC平分线交AC于D,
CE是∠ACB平分线交AB于E,BD,CE交于P,
设∠BPC＞90°,
∠BPC=180°-1/2∠B-1/2∠C,
=90°+1/2（∠A+∠B+∠C）-1/2∠B-1/2∠C
=90°+1/2∠A.
3.设△ABC中,BD是∠ABC的平分线,
CD是∠ACB外角平分线,
BD,CD交于D,
∴∠BDC=180°-1/2∠B-∠C-1/2（∠A+∠B）
=180°-1/2∠B-∠C-1/2∠A-1/2∠B
=180°-∠B-∠C-1/2∠A
=180°-∠B-∠C-∠A+∠A-1/2∠A
=1/2∠A.
证毕.