已知:如图梯形ABCD中,AD‖BC,AB=CD,P为BC上一点,∠APQ=∠B,PQ交CD于Q 求证:△ABP∽△PCQ
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![已知:如图梯形ABCD中,AD‖BC,AB=CD,P为BC上一点,∠APQ=∠B,PQ交CD于Q 求证:△ABP∽△PCQ](/uploads/image/z/13948159-31-9.jpg?t=%E5%B7%B2%E7%9F%A5%3A%E5%A6%82%E5%9B%BE%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%80%96BC%2CAB%3DCD%2CP%E4%B8%BABC%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E2%88%A0APQ%3D%E2%88%A0B%2CPQ%E4%BA%A4CD%E4%BA%8EQ+%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3ABP%E2%88%BD%E2%96%B3PCQ)
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