Question

如图所示,在梯形ABCD中,AD//BC,AP平分∠BAD,BP平分∠ABC,AP、BP交于点P,求∠APB的度数.

Answer 1

图形样子大致就那样的吧

∵AD∥BC

∴∠BAD+∠ABC=180

又∵AP平分∠BAD,BP平分∠ABC

∴∠PAB=1/2∠BAD,∠PBA=1/2∠ABC

∴∠PAB+∠PBA=1/2∠BAD+1/2∠ABC=1/2（∠BAD+∠ABC）=1/2×180=90

∵∠PAB+∠PBA+∠APB=180

∴∠APB=90