题目如图,已知△ABC\triangle ABC△ABC中,AB=2AB=2AB=2,AC=3AC=3AC=3,AD⊥BCAD\bot BCAD⊥BC于DDD,PPP为ADADAD上任一点,则PC2−PB2PC^{2}-PB^{2}PC2−PB2等于( )A.555B.666C.777D.888知识点:三角形边角关系章节:未标注