题目在正方形ABCD中,P为BC上任意一点,PQ⊥AP且交∠BCD的外角的平分线于Q。求证:AP=PQ。(图1) 证明连接A、C,A、Q,则A、P、C、Q四点共圆。∴∠AQP=∠ACP=45°,AP=PQ。这是一个耐人寻味的几何题,它可以有多种证法,还可以把它推广到更为一般的情形。在正n边形AA_1 A_2…A_(n-1)中,P是边A_1A_2上任意一点,作∠APQ=∠A_1,PQ交∠A_1A_2A_3的外角∠A_3A_2M的平分线于Q,则AP=PQ。(图2) 证明连接A、A_2,A、Q。∵∠APQ=∠A_1=(n-2)180°/n,AA_1=A_1A_2, In the square ABCD, P is an arbitrary point on BC, and the bisector of the outer corner of PQ⊥AP and BCD is Q. Proof: AP=PQ. (Figure 1) Prove that connect A, C, A, Q, A, P, C, Q four points are co-circular. ∴∠ AQP=∠ACP=45°, AP=PQ. This is an intriguing geometry. It can have multiple proofs and it can be extended to more general situations. In the positive-n-angle AA_1 A_2...A_(n-1), P is an arbitrary point on the edge A_1A_2, for ∠APQ=∠A_1, the bisector of the external angle ∠A_3A_2M of PQ intersects A_1A_2A_3 is Q, and then AP=PQ. (Figure 2) Proof connection A, A_2, A, Q. ∵∠APQ=∠A_1=(n-2)180°/n,AA_1=A_1A_2,