ABCD is a square, ABE and CMN are equilateral triangles inside the square, E onto CM. AE with MN intersect at Q, BE with NC intersect at P. Prove that PQ_|_CN.
Proof:
<DCM=<BCN=15 (to get <MCN=60 and CM=CN), thus <AMN=<ANM=45. Next ANEM and CENB are cyclic, so AE is angle bisector of <MEN=90, <NEB=<NCB=15, <CNE=<CBE=30 and easy angles calculations give <MQE=75, <CPE=45.
As <PEQ=(<NQP+<NPQ)/2=60, E , onto the bisector of <MNC is the N-excenter of tr. NPQ, hence <EPQ=<EPC=45, consequently <CPQ=90, done.
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