Because MMP is goated.
A point \(P\) inside a triangle \(ABC\) is given. The lines \(BP\), \(CP\) meet the circle \(ABC\) for the second time at points \(E\), \(F\) respectively. The circle \(\Omega\) passes through \(P\), \(E\) and meets \(AC\) at points \(B_1\), \(B_2\). The lines \(PB_1\), \(PB_2\) meet \(AB\) at points \(C_1\), \(C_2\). Prove that \(C_1\), \(C_2\), \(P\), \(F\) are concyclic.
\(P\) is fixed. \[B_1 \; [AC] \to^? B_2 \; [AC] \to PB_2 \cap AB = C_2 \; [AB]\] \[B_1 \; [AC] \to B_1 P \cap AB = C_1 \; [AB] \to^? C_2' \; [AB]\] But inversion at \(BP \cap AC = K\) with radius \(KP \cdot KE\) is a fixed inversion that sends \(AC \to AC\) and \(B_1 \to B_2\). Similarly, the fixed inversion at \(CP \cap AB = L\) with radius \(LP \cdot LF\) sends \(AB \to AB\) and \(C_1 \to C_2'\). Thus the \(\to^?\) are actually projective.
Therefore, \(C_2 \equiv C_2'\) is a projective condition and we only need to check \(3\) cases of \(B_1\) (blah blah blah)
Let \(ABC\) be a triangle with \(\angle A = 120^{\circ}\); \(P\) be an arbitrary point inside this triangle lying on the bisector of angle \(A\); the lines \(BP\), \(CP\) meet \(AC\), \(AB\) at points \(E\), \(F\) respectively; \(D\) be an arbitrary point on the side \(BC\); the lines \(DE\), \(DF\) meet \(PC\), \(PB\) at points \(M\), \(N\) respectivly. Find the value of angle \(MAN\).
\(P\) is fixed. We want to show some two definitions of \(M\) are equivalent to prove \(\angle MAN\) is fixed (we claim that it is \(60^{\circ}\)). The first definition is simple: \[ D \; [BC] \to M \; [CP] \] Take \(B'P\), the fixed reflection line of \(BP\) about \(AP\). Then \[D \; [BC] \to N \; [BP] \to N' \; [B'P] \;\;\;\; (\text{projection through Infinity perpendicular to } AP)\] \[N' \; [B'P] \to X \; [BP] \;\;\;\; (\text{projection through } A)\] Thus we have projectively achieved \(X\), the Intersection(Reflect(\(AN\), \(AP\)), \(BP\)). Through the same procedure, reflecting about angle bisector of \(\angle CAP\), we can go from \(Y\) to \(M\) where \(Y\) is: \[X \; [BP] \to Y \; [CP] \;\;\;\; (\text{projection through } A)\] \[Y \; [CP] \to M \; [CP] \;\;\;\; (\text{procedure}) \] Therefore, since both these maps from \(D\) to \(M\) are projective, to show their congruence, we need to check \(3\) cases of \(D\) (blah blah blah)
Instead of taking a new line and projecting through an Infinity, note that rotation about a fixed point through a fixed angle is literally a projective map, so we can just directly rotate by \(60^{\circ}\) instead and get the same result, much faster. Also, a reflection through a fixed line is a projective map, so we were just completely coping, because we could have just directly taken the reflections about the angle bisectors.
Let \(P\), \(Q\) be isogonal conjugate points wrt. \(\triangle ABC\). Point \(X\) is the orthogonal projection of \(Q\) on \(BC\). The circle with diameter \(PA\) intersects \((ABC)\) at \(K \neq A\). \(AQ\) intersects \((ABC)\) at \(T \neq A\). Prove that the points \(K\), \(X\), \(T\) are collinear.
Let \(Q\) be tethered on \(AT\), and \(P\) be tethered on the isogonal line of \(AT\), say \(AS\). Note that \[Q \; [AT] \to X \; [BC] \to K \; [(ABC)]\] is projective, and \[Q \; [AT] \to BQ \; [\mathbf{C}_B] \to BP \; [\mathbf{C}_B] \to P \; [AS]\] is projective. So we want to show that \(P \; [AS] \to K' \; [(ABC)]\) is projective, where \(K'\) is the intersection of \((ABC)\) and the circle with diameter \(AP\). But of course, introducing \(A'\), the antipode of \(A\) in \((ABC)\), note that \(A'\), \(P\), \(K'\) are collinear, so projecting through \(A'\), indeed \(P \; [AS] \to K' \; [(ABC)]\) is projective.
Therefore we have a projective map from \(Q \to K\) and \(Q \to K'\), and to show their equivalence we must check \(3\) cases for \(Q\) (blah blah blah)
Alternatively, to obtain \(P \; [AS] \to K' \; [(ABC)]\) is projective, we can invert at \(A\) and show that the inverted \(P \to K'\) is projective, so that \(P \to K'\) in the non-inverted problem is also projective (inversion preserves cross-ratios).
Convex pentagon \(AXBCY\) is inscribed in circle \(\Omega\). Suppose diagonal \(XY\) meets segments \(AB\) and \(AC\) at \(D\) and \(E\) respectively. Show that the midpoints of \(BE\), \(CD\), \(XY\), and \(DE\) are concyclic.
In \( \triangle ABC \), \( \angle B \) is obtuse and \( AB \neq BC \). Let \( O \) be the circumcenter and \( \omega \) be the circumcircle of this triangle. \( N \) is the midpoint of arc \( ABC \). The circumcircle of \( \triangle BON \) intersects \( AC \) on points \( X \) and \( Y \). Let \( BX \cap \omega = P \neq B \) and \( BY \cap \omega = Q \neq B \). Prove that \( P\), \(Q \) and reflection of \( N \) with respect to line \( AC \) are collinear.
Given a \( \triangle ABC \) and two isogonal points \( P \) and \( Q \). Points \( D\), \(E \) and \( F \) are intersections of lines \( AP\), \(BP \) and \( CP \) with sides \( BC\), \(AC \), and \( AB \), respectively. Let \( O \) be the circumcenter of \( \triangle ABC \). Let \( X \) be the intersection of the line perpendicular to \( EF \) through \( A \) and the line \( OD \). Prove that \( QX \perp BC \).
Let \(ABCD\) be a convex quadrilateral satisfying \(\angle BAD + 2\angle BCD = 180^{\circ}\). Let \(E\) be the intersection of \(BD\) and the internal bisector of \(\angle BAD\). The perpendicular bisector of \(AE\) intersects lines \(CB\) and \(CD\) at \(X\) and \(Y\) respectively. Prove that \(A\), \(X\), \(C\), \(Y\) are concyclic.
Let \(ABC\) be a non-equilateral triangle and let \(M_a\), \(M_b\), \(M_c\) be the midpoints of the sides \(BC\), \(CA\), \(AB\), respectively. Let \(S\) be a point lying on the Euler line. Denote by \(X\), \(Y\), \(Z\) the second intersections of \(M_a S\), \(M_b S\), \(M_c S\) with the nine-point circle. Prove that \(AX\), \(BY\), \(CZ\) are concurrent.
Two circles \(\Gamma_1\) and \(\Gamma_2\) have common external tangents \(\ell_1\) and \(\ell_2\) meeting at \(T\). Suppose \(\ell_1\) touches \(\Gamma_1\) at \(A\) and \(\ell_2\) touches \(\Gamma_2\) at \(B\). A circle \(\Omega\) through \(A\) and \(B\) intersects \(\Gamma_1\) again at \(C\) and \(\Gamma_2\) again at \(D\), such that quadrilateral \(ABCD\) is convex. Suppose lines \(AC\) and \(BD\) meet at point \(X\) while lines \(AD\) and \(BC\) meet at point \(Y\). Show that \(T\), \(X\), \(Y\) are collinear.
Let \(ABC\) be a triangle and let \(M\) and \(N\) denote the midpoints of \({AB}\) and \({AC}\), respectively. Let \(X\) be a point such that \({AX}\) is tangent to the circumcircle of triangle \(ABC\). Denote by \(\omega_B\) the circle through \(M\) and \(B\) tangent to \({MX}\), and by \(\omega_C\) the circle through \(N\) and \(C\) tangent to \({NX}\). Show that \(\omega_B\) and \(\omega_C\) intersect on line \(BC\).
Let \(ABC\) be a fixed triangle with circumcircle \(\gamma\), and let \(P\) be any point in its interior. Ray \(AP\) meets \(\gamma\) again at \(A_1\). We reflect \(A_1\) across \(BC\) to obtain a point \(A_2\). Define \(B_1\), \(B_2\), \(C_1\) and \(C_2\) similarly. Prove that the circumcircle of \(A_2B_2C_2\) passes through a fixed point independent of \(P\).
Let \(ABC\) be an acute triangle with orthocenter \(H\) and circumcircle \(\Gamma\). A line through \(H\) intersects segments \(AB\) and \(AC\) at \(E\) and \(F\), respectively. Let \(K\) be the circumcenter of \(\triangle AEF\), and suppose line \(AK\) intersects \(\Gamma\) again at a point \(D\). Prove that line \(HK\) and the line through \(D\) perpendicular to \({BC}\) meet on \(\Gamma\).
Let \(ABC\) be an acute triangle with circumcircle \(\Omega\) and orthocenter \(H\). Points \(D\) and \(E\) lie on segments \(AB\) and \(AC\) respectively, such that \(AD = AE\). The lines through \(B\) and \(C\) parallel to \(\overline{DE}\) intersect \(\Omega\) again at \(P\) and \(Q\), respectively. Denote by \(\omega\) the circumcircle of \(\triangle ADE\).
Let \(ABC\) be a triangle with circumcircle \(\omega\) and \(\ell\) a line without common points with \(\omega\). Denote by \(P\) the foot of the perpendicular from the center of \(\omega\) to \(\ell\). The side-lines \(BC\), \(CA\), \(AB\) intersect \(\ell\) at the points \(X\), \(Y\), \(Z\) different from \(P\). Prove that the circumcircles of the triangles \(AXP\), \(BYP\) and \(CZP\) have a common point different from \(P\) or are mutually tangent at \(P\).