In acute △ABC, K is on the extention of segment BC. P,Q are two points such that KP∥AB,BK=BP and KQ∥AC,CK=CQ. The circumcircle of △KPQ intersects AK again at T. Prove that:
(1) ∠BTC+∠APB=∠CQA.
(2) AP⋅BT⋅CQ=AQ⋅CT⋅BP.Proposed by Yijie He and Yijuan Yao