Question

三角形的外角的练习如图,BP平分∠ABC交于CD于F,DP平分∠ADC交于AB于E,若∠A=38°,∠C=46°,求∠P

Answer 1

∠P=∠ABF-∠BEP

=∠AEC/2-(180-∠A-∠ADE)

=∠AEC/2-(180-∠A-∠ADC/2)

=(∠AEC+∠ADC)/2+∠A-180

=(360-∠A-∠C)/2+38-180

=(360-38-46)/2-142

=-4

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∠P=180-∠FBC-∠BEP

=180-∠ABC/2-(∠EDC+∠C)

=180-∠ABC/2-∠ADC/2-∠C

=(360-(∠ABC+∠ADC))/2-∠C

=(∠A+∠C)/2-∠C

=(∠A- ∠C)/2

=-4

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交点在ABCD内外都是-4度

题目有问题,BP应该不与CD相交,或者是交于延长线