Let ABC be a triangle and P, Q points onto the side BC so that <BAP=<ACB, <CAQ=<ABC; on the lines AP, AQ take points M,N as symmetricals of A about P, Q respectively and call L interesection of BM with CN. Prove that ABLC is cyclic.
Proof.
As constructed, both triangles ABP, CAQ are similar to CBA, and subsequently they are similar, AP=AQ and AP/CQ=BP/AQ ( 1 ). But AP=MP, QA=QN, so (1) becomes MP/CQ=BP/QN, which means triangles BPM and NQC are similar (<BPM=<CQN also), from their similarity getting <QCN=<BMP, i.e. BLNQ, CPML are cyclic, so <MLN=<BPM=<APQ=180-<BAC, done.
NOTE: As <AMN=<ANM=<MLN, AL is symmedian of triangle MLN, thus of triangle BLC as well and ABLC is harmonic quadrilateral.
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