practice problems
Q1. (ABC) is a triangle. (D, E, F) are arbitrary points on (BC, AC,) and (AB) respectively (or on their extensions). Draw the three circumcircles of triangles (AEF), (DBF), and (DEC). Prove that these three circles intersect at a single point (M). S1. This is known as Miquel's theorem and M is known as Miquel's point. Case 1. When AEF,DBF intersect at 2 different points : F and M. Proof 1: Angle AFM = theta => MFB = 180 - theta (supplementary angles) and AEM = 180 - theta (opposite angles in cyclic quadrilateral AEMF) => MEC = theta (supplementary) and BDM = theta (opposite in cyclic quadrilateral BDMF) => CDM = 180 - theta (supplementary) => CEM + CDM = 180 => CEMD is cyclic quadrilateral H.P. Proof 2: AFME is cyclic => FME = 180 - A = B + C Similarly, FMD = A + B Now angles around M should add up to 360. => FME + EMD + DMF = 360 => EMD = 360 - (B+C) - (A+B) = So we showed that M is the Miquel's point. But could it have been F? I mean, is it possib...