Question

如图，已知：在Rt△ABC中，∠C=90°，BD平分∠ABC且交AC于D.若AP平分∠BAC且交BD于点P,求∠BPA的度数

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Answer 1

∠ABP=1/2∠ABC
∠BAP=1/2∠BAC
∠BPA=180-∠ABP-∠BAP=180-1/2∠ABC-1/2∠BAC=180-1/2(∠ABC+∠BAC)
因为∠ABC+∠BAC+∠C=180
∠C==90
所以∠ABC+∠BAC=90
所以∠BPA=180-1/2(∠ABC+∠BAC)=180-45=135

Answer 2

∠ABP=1/2∠ABC（BD平分∠ABC）
∠BAP=1/2∠BAC（AP平分∠BAC）
又∵∠ABC+∠BAC=180°-∠C=180°-90°=90°

∠ABP+∠BAP=1/2(∠ABC+∠BAC)=45°
∴∠BPA=180°-（∠ABP+∠BAP）=135°

Answer 3

∠BPA=135°
∠BPA=180°-(∠ABP-∠BAP)=180°-1/2(∠ABC+∠BAC)=180°-45°=135°

Answer 4

角BPA=180-(角ABP+角BAP)=180-1/2(角ABC+角BAC)=180-1/2*90=180-45=135