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如图,已知点 P 是矩形 ABCD 内一点(不含边界),设 ∠ PAD = θ 1 , ∠ PBA = θ 2 , ∠ PCB = θ 3 , ∠ PDC = θ 4 ,若 ∠ APB = 80 ° , ∠ CPD = 50 ° ,则 ( )
A. ( θ 1 + θ 4 ) − ( θ 2 + θ 3 ) = 30 ° B. ( θ 2 + θ 4 ) − ( θ 1 + θ 3 ) = 40 °
C. ( θ 1 + θ 2 ) − ( θ 3 + θ 4 ) = 70 ° D. ( θ 1 + θ 2 ) + ( θ 3 + θ 4 ) = 180 °
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