diff --git a/project_q1/Basic.lean b/project_q1/Basic.lean
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+-- 五点共圆（JZM问题）
+-- 林元莘 王孙葳 上海交通大学
+
+import EuclideanGeometry.Foundation.Index
+
+noncomputable section
+
+namespace EuclidGeom
+
+variable {Plane : Type _} [EuclideanPlane Plane]
+
+-- There are temporarily no functions for the circumcircle of three points and the perpendicular bisector before the completion of this section of the code.
+-- We dicide to creat one.
+
+-- Perpendicular bisector
+
+def perp_bis (A B : Plane)(h: B ≠ A) : Line Plane :=
+  perp_line ((SEG A B).midpoint) (LIN A B h)
+
+-- The points on the perpendicular bisector of a line segment are equidistant from the two endpoints.
+-- We haven't complete that yet.
+theorem dist_eq_of_on_perp_bis {A B C : Plane} (h: B ≠ A) (h1: C LiesOn perp_bis A B h) :
+   dist A C = dist B C := by sorry
+
+-- The circumcenter of a triangle is the intersection point of the perpendicular bisectors of any two sides.
+
+def circumcenter (tr : TriangleND Plane) : Plane :=
+-- Annoying collinearity
+  have tr_bis_intersect : ¬ (perp_bis tr.1.1 tr.1.2 (ne_of_not_collinear tr.2).2.2) ∥ (perp_bis tr.1.2 tr.1.3 (ne_of_not_collinear tr.2).1) := sorry
+  Line.inx (perp_bis tr.1.1 tr.1.2 (ne_of_not_collinear tr.2).2.2) (perp_bis tr.1.2 tr.1.3 (ne_of_not_collinear tr.2).1) tr_bis_intersect
+
+-- The circumcenter of a triangle is equidistant from the three vertices.
+theorem circumcenter_dist_vetex_eq {tr : TriangleND Plane}: dist tr.1.1 (circumcenter tr) = dist tr.1.2 (circumcenter tr) ∧ dist tr.1.2 (circumcenter tr) = dist tr.1.3 (circumcenter tr) ∧ dist tr.1.1 (circumcenter tr) = dist tr.1.3 (circumcenter tr) := by
+
+  have O1_eq_O2 : dist tr.1.1 (circumcenter tr) = dist tr.1.2 (circumcenter tr) := by
+    apply dist_eq_of_on_perp_bis (ne_of_not_collinear tr.2).2.2 _
+    have tr_bis_intersect : ¬ (perp_bis tr.1.1 tr.1.2 (ne_of_not_collinear tr.2).2.2) ∥ (perp_bis tr.1.2 tr.1.3 (ne_of_not_collinear tr.2).1) := sorry
+    rw[circumcenter]
+    apply Line.inx_lies_on_fst tr_bis_intersect
+  have O2_eq_O3 : dist tr.1.2 (circumcenter tr) = dist tr.1.3 (circumcenter tr) := by
+    apply dist_eq_of_on_perp_bis (ne_of_not_collinear tr.2).1 _
+    have tr_bis_intersect : ¬ (perp_bis tr.1.1 tr.1.2 (ne_of_not_collinear tr.2).2.2) ∥ (perp_bis tr.1.2 tr.1.3 (ne_of_not_collinear tr.2).1) := sorry
+    rw[circumcenter]
+    apply Line.inx_lies_on_snd tr_bis_intersect
+  constructor
+  exact O1_eq_O2
+  constructor
+  exact O2_eq_O3
+  rw[O1_eq_O2, O2_eq_O3]
+
+-- Annoying Non-zero
+theorem circum_not_vertex {tr : TriangleND Plane}: 0 < dist (circumcenter tr) tr.1.1 ∧ 0 < dist tr.1.2 (circumcenter tr)  ∧0 <  dist tr.1.3 (circumcenter tr) := by
+  sorry
+
+
+-- Three non-collinear points determine a circle.
+
+def mk_pt_pt_pt_circle (A B C : Plane) (h: ¬Collinear A B C) : Circle Plane where
+  center := circumcenter (TRI_nd A B C h)
+  radius := dist (circumcenter (TRI_nd A B C h)) A
+  rad_pos := by
+    let tr := TRI_nd A B C h
+    have h₁ : 0 < dist (circumcenter tr) tr.1.1 := by exact (circum_not_vertex).1
+    exact h₁
+
+-- Several useful but annoying theorems
+
+
+
+-- Points lies on the circle we made.
+theorem pt_on_mk_pt_pt_pt_circle {A B C : Plane} (h: ¬Collinear A B C) : A LiesOn mk_pt_pt_pt_circle A B C h ∧ B LiesOn mk_pt_pt_pt_circle A B C h ∧ C LiesOn mk_pt_pt_pt_circle A B C h :=by
+  constructor
+  rfl
+  constructor
+  have h₁ : dist A (circumcenter (TRI_nd A B C h)) = dist B (circumcenter (TRI_nd A B C h)) := by
+    have q₁ : A = (TRI_nd A B C h).1.1 := by rfl
+    have q₂ : B = (TRI_nd A B C h).1.2 := by rfl
+    nth_rw 1 [q₁]
+    nth_rw 3 [q₂]
+    sorry
+  sorry
+  sorry
+
+
+
+-- Criterion for four points to be concyclic
+theorem four_pt_circle {A B C D : Plane} (h₁ : ¬Collinear A B C) (h₂ : A ≠ D) (h₃ : C ≠ D) (hang : (ANG A B C (ne_of_not_collinear h₁).2.2.symm  (ne_of_not_collinear h₁).1).dvalue = (ANG A D C h₂ h₃).dvalue) : D LiesOn mk_pt_pt_pt_circle A B C h₁ := by
+  sorry
+-- One side of the two angles is collinear.
+theorem dvalue_eq_dir1_one_line {A B O Q : Plane} (l : Line Plane) (h : A LiesOn l ∧ B LiesOn l ∧ O LiesOn l)(h₁: ¬ Collinear A O Q) (h₂ : ¬ Collinear B O Q) : (ANG A O Q (ne_of_not_collinear h₁).2.2.symm  (ne_of_not_collinear h₁).1).dvalue = (ANG B O Q (ne_of_not_collinear h₂).2.2.symm  (ne_of_not_collinear h₂).1).dvalue := by
+  sorry
+
+theorem dvalue_eq_dir2_one_line {A B O Q : Plane} (l : Line Plane) (h : A LiesOn l ∧ B LiesOn l ∧ O LiesOn l)(h₁: ¬ Collinear Q O A) (h₂ : ¬ Collinear Q O B) : (ANG Q O A (ne_of_not_collinear h₁).2.2.symm  (ne_of_not_collinear h₁).1).dvalue = (ANG Q O B (ne_of_not_collinear h₂).2.2.symm  (ne_of_not_collinear h₂).1).dvalue := by
+  sorry
+
+-- Inscribed angles are equal modulo π.
+theorem iangle_theorem {A B C D : Plane} (O : Circle Plane) (h₁ : ¬ Collinear A B C) (h₂ : ¬ Collinear A D C) (hc : A LiesOn O ∧ B LiesOn O ∧ C LiesOn O ∧ D LiesOn O) : (ANG A B C (ne_of_not_collinear h₁).2.2.symm  (ne_of_not_collinear h₁).1).dvalue = (ANG A D C (ne_of_not_collinear h₂).2.2.symm  (ne_of_not_collinear h₂).1).dvalue := by
+  sorry
+
+-- addition for angles with a common vertex.
+theorem dvalue_add_theorem {A B C O : Plane} (h₁ : ¬ Collinear A O B) (h₂ : ¬ Collinear A O C) (h₃ : ¬ Collinear C O B) :
+(ANG A O B (ne_of_not_collinear h₁).2.2.symm (ne_of_not_collinear h₁).1).dvalue = (ANG A O C (ne_of_not_collinear h₂).2.2.symm (ne_of_not_collinear h₂).1).dvalue + (ANG C O B (ne_of_not_collinear h₃).2.2.symm (ne_of_not_collinear h₃).1).dvalue := by
+sorry
+
+-- Now we start creating the STAR.
+structure Setting (Plane : Type _) [EuclideanPlane Plane] where
+-- Firstly, we creat 5 points to generate a Pentagon ABCDE.
+  A : Plane
+  B : Plane
+  C : Plane
+  D : Plane
+  E : Plane
+  B_ne_A : B ≠ A
+  C_ne_B : C ≠ B
+  D_ne_C : D ≠ C
+  E_ne_D : E ≠ D
+  A_ne_E : A ≠ E
+-- These ensure that the pentagon is non-degenerate.
+  hp₁ : ¬ Collinear A B C
+  hp₂ : ¬ Collinear B C D
+  hp₃ : ¬ Collinear C D E
+  hp₄ : ¬ Collinear D E A
+  hp₅ : ¬ Collinear E A B
+  l₁ := LIN A B B_ne_A
+  l₂ := LIN B C C_ne_B
+  l₃ := LIN C D D_ne_C
+  l₄ := LIN D E E_ne_D
+  l₅ := LIN E A A_ne_E
+-- Non-parallelism ensures the creation of a STAR FGHIJ-ABCDE.
+  h₁ : ¬ l₁ ∥ l₃
+  h₂ : ¬ l₂ ∥ l₄
+  h₃ : ¬ l₃ ∥ l₅
+  h₄ : ¬ l₄ ∥ l₁
+  h₅ : ¬ l₅ ∥ l₂
+  F : Plane
+  G : Plane
+  H : Plane
+  I : Plane
+  J : Plane
+  F_inx : is_inx F l₁ l₃
+  G_inx : is_inx G l₂ l₄
+  H_inx : is_inx H l₃ l₅
+  I_inx : is_inx I l₄ l₁
+  J_inx : is_inx J l₅ l₂
+-- Some conditions for laziness
+  IBF_on_l₁ : I LiesOn l₁ ∧ B LiesOn l₁ ∧ F LiesOn l₁ := by sorry
+  IBA_on_l₁ : I LiesOn l₁ ∧ B LiesOn l₁ ∧ A LiesOn l₁ := by sorry
+  BGC_on_l₂ : B LiesOn l₂ ∧ G LiesOn l₂ ∧ C LiesOn l₂ := by sorry
+  GID_on_l₄ : G LiesOn l₄ ∧ I LiesOn l₄ ∧ D LiesOn l₄ := by sorry
+  hF : ¬ Collinear B C F := by sorry
+  hG : ¬ Collinear C D G := by sorry
+  hH : ¬ Collinear D E H := by sorry
+  hI : ¬ Collinear E A I := by sorry
+  hJ : ¬ Collinear A B J := by sorry
+  tr₁ := TRI_nd B C F hF
+  tr₂ := TRI_nd C D G hG
+  tr₃ := TRI_nd D E H hH
+  tr₄ := TRI_nd E A I hI
+  tr₅ := TRI_nd A B J hJ
+-- Creat 5 circles
+  C₁ : Circle Plane
+  hc1 : C₁ = mk_pt_pt_pt_circle B C F hF
+  C₂ : Circle Plane
+  hc2 : C₂ = mk_pt_pt_pt_circle C D G hG
+  C₃ : Circle Plane
+  hc3 : C₃ = mk_pt_pt_pt_circle D E H hH
+  C₄ : Circle Plane
+  hc4 : C₄ = mk_pt_pt_pt_circle E A I hI
+  C₅ : Circle Plane
+  hc5 : C₅ = mk_pt_pt_pt_circle A B J hJ
+  c₁intc₂ : C₁ Intersect C₂
+  c₂intc₃ : C₂ Intersect C₃
+  c₃intc₄ : C₃ Intersect C₄
+  c₄intc₅ : C₄ Intersect C₅
+  c₅intc₁ : C₅ Intersect C₁
+  K : Plane
+  hK : K LiesOn C₅ ∧ K LiesOn C₁ ∧ K ≠ B
+  L : Plane
+  hL : L LiesOn C₁ ∧ L LiesOn C₂ ∧ L ≠ C
+  M : Plane
+  hM : M LiesOn C₂ ∧ M LiesOn C₃ ∧ M ≠ D
+  N : Plane
+  hN : N LiesOn C₃ ∧ N LiesOn C₄ ∧ N ≠ E
+  O : Plane
+  hP : O LiesOn C₄ ∧ O LiesOn C₅ ∧ O ≠ A
+-- Some conditions for laziness
+  BFLC_on_C₁ : B LiesOn C₁ ∧ F LiesOn C₁ ∧ L LiesOn C₁ ∧ C LiesOn C₁ := by sorry
+  LKBF_on_C₁ : L LiesOn C₁ ∧ K LiesOn C₁ ∧ B LiesOn C₁ ∧ F LiesOn C₁ := by sorry
+  GDLC_on_C₂ : G LiesOn C₂ ∧ D LiesOn C₂ ∧ L LiesOn C₂ ∧ C LiesOn C₂ := by sorry
+  INOA_on_C₄ : I LiesOn C₄ ∧ N LiesOn C₄ ∧ O LiesOn C₄ ∧ A LiesOn C₄ := by sorry
+  BKOA_on_C₅ : B LiesOn C₅ ∧ K LiesOn C₅ ∧ O LiesOn C₅ ∧ A LiesOn C₅ := by sorry
+  IFL : ¬ Collinear I F L := by sorry
+  LKO : ¬ Collinear L K O := by sorry
+  IDL : ¬ Collinear I D L := by sorry
+  BFL : ¬ Collinear B F L := by sorry
+  BCL : ¬ Collinear B C L := by sorry
+  GDL : ¬ Collinear G D L := by sorry
+  LKB : ¬ Collinear L K B := by sorry
+  BKO : ¬ Collinear B K O := by sorry
+  LFI : ¬ Collinear L F I := by sorry
+  BAO : ¬ Collinear B A O := by sorry
+  LNO : ¬ Collinear L N O := by sorry
+  LNI : ¬ Collinear L N I := by sorry
+  INO : ¬ Collinear I N O := by sorry
+  GCL : ¬ Collinear G C L := by sorry
+  LFB : ¬ Collinear L F B := by sorry
+  IAO : ¬ Collinear I A O := by sorry
+
+-- Restate our problem : A STAR FGHIJ-ABCDE, create circumcircles of 5 triangles : C₁ for ▵BCF, C₂ for ▵CDG, C₃ for ▵DHE,
+-- C₄ for ▵EAI, C₅ for ▵ABJ. Two adjacent circles intersect at two different points : K,B ∈ C₅ ∩ C₁, L,C ∈ C₁ ∩ C₂, M,D ∈ C₂ ∩ C₃, N,E ∈ C₃ ∩ C₄, O,A ∈ C₄ ∩ C₅
+-- Prove: K,L,M,N,O are concyclic
+
+theorem JZM_Theorem {Plane : Type _} [EuclideanPlane Plane] (e : Setting Plane): ∃ (X : Circle Plane), e.K LiesOn X ∧ e.L LiesOn X ∧ e.M LiesOn X ∧ e.N LiesOn X ∧ e.O LiesOn X := by
+-- First, we prove that I,F,L,D are concyclic
+  have IFLD_c : e.D LiesOn mk_pt_pt_pt_circle e.I e.F e.L e.IFL := by
+  -- Prove ∠IFL = ∠GDL (mod π) (the equalities below are under the meaning of modulo π)
+    have IFL_eq_GDL : (ANG e.I e.F e.L (ne_of_not_collinear e.IFL).2.2.symm  (ne_of_not_collinear e.IFL).1).dvalue = (ANG e.I e.D e.L (ne_of_not_collinear e.IDL).2.2.symm  (ne_of_not_collinear e.IDL).1).dvalue := by
+    -- We can prove it by ∠IFL = ∠BCL = ∠GCL = ∠GDL
+      -- Prove ∠IFL = ∠BFL by collinearity.
+      have s₁ : (ANG e.I e.F e.L (ne_of_not_collinear e.IFL).2.2.symm  (ne_of_not_collinear e.IFL).1).dvalue = (ANG e.B e.F e.L (ne_of_not_collinear e.BFL).2.2.symm  (ne_of_not_collinear e.BFL).1).dvalue := by
+        exact dvalue_eq_dir1_one_line e.l₁ e.IBF_on_l₁ e.IFL e.BFL
+      -- Prove ∠BFL = ∠BCL by concyclicity
+      have s₂ : (ANG e.B e.F e.L (ne_of_not_collinear e.BFL).2.2.symm  (ne_of_not_collinear e.BFL).1).dvalue = (ANG e.B e.C e.L (ne_of_not_collinear e.BCL).2.2.symm  (ne_of_not_collinear e.BCL).1).dvalue := by
+        exact iangle_theorem e.C₁ e.BFL e.BCL e.BFLC_on_C₁
+      -- Prove ∠BCL = ∠ GCL by collinearity
+      have s₃ : (ANG e.B e.C e.L (ne_of_not_collinear e.BCL).2.2.symm  (ne_of_not_collinear e.BCL).1).dvalue = (ANG e.G e.D e.L (ne_of_not_collinear e.GDL).2.2.symm  (ne_of_not_collinear e.GDL).1).dvalue := by
+        rw [iangle_theorem e.C₂ e.GDL e.GCL e.GDLC_on_C₂]
+        exact dvalue_eq_dir1_one_line e.l₂ e.BGC_on_l₂ e.BCL e.GCL
+      rw[s₁,s₂,s₃]
+      exact dvalue_eq_dir1_one_line e.l₄ e.GID_on_l₄ e.GDL e.IDL
+    exact four_pt_circle (e.IFL) (ne_of_not_collinear e.IDL).2.2.symm (ne_of_not_collinear e.IDL).1 IFL_eq_GDL
+  -- Prove I,F,N,L are concyclic. Only need to prove ∠FIN = ∠FDN as above.
+  have IFNL_c : e.N LiesOn mk_pt_pt_pt_circle e.I e.F e.L e.IFL := by sorry
+-- Second, we have I,F,N,L,D are concyclic. We can prove that K,L,O,N are concyclic.
+  have KLON_c : e.N LiesOn mk_pt_pt_pt_circle e.L e.K e.O e.LKO := by
+    let aLFI := (ANG e.L e.F e.I (ne_of_not_collinear e.LFI).2.2.symm  (ne_of_not_collinear e.LFI).1)
+    let aBAO := (ANG e.B e.A e.O (ne_of_not_collinear e.BAO).2.2.symm  (ne_of_not_collinear e.BAO).1)
+    let aLNO := (ANG e.L e.N e.O (ne_of_not_collinear e.LNO).2.2.symm  (ne_of_not_collinear e.LNO).1)
+    let aLNI := (ANG e.L e.N e.I (ne_of_not_collinear e.LNI).2.2.symm  (ne_of_not_collinear e.LNI).1)
+    let aINO := (ANG e.I e.N e.O (ne_of_not_collinear e.INO).2.2.symm  (ne_of_not_collinear e.INO).1)
+  -- Prove ∠LKO = ∠LNO
+    -- Prove ∠LKO = ∠LKB + ∠BKO = ∠LFI + ∠BAO
+    have LKO_eq_LFI_ad_BAO : (ANG e.L e.K e.O (ne_of_not_collinear e.LKO).2.2.symm  (ne_of_not_collinear e.LKO).1).dvalue = (ANG e.L e.F e.I (ne_of_not_collinear e.LFI).2.2.symm  (ne_of_not_collinear e.LFI).1).dvalue + (ANG e.B e.A e.O (ne_of_not_collinear e.BAO).2.2.symm  (ne_of_not_collinear e.BAO).1).dvalue := by
+      -- Prove ∠LKB = ∠LFI
+      have LKB_eq_LFI : (ANG e.L e.K e.B (ne_of_not_collinear e.LKB).2.2.symm  (ne_of_not_collinear e.LKB).1).dvalue = (ANG e.L e.F e.I (ne_of_not_collinear e.LFI).2.2.symm  (ne_of_not_collinear e.LFI).1).dvalue := by
+        rw[dvalue_eq_dir2_one_line e.l₁ e.IBF_on_l₁ e.LFI e.LFB] --∠LFI = ∠LFB
+        exact iangle_theorem e.C₁ e.LKB e.LFB e.LKBF_on_C₁ -- ∠LKB = ∠LFB
+      -- Prove ∠BKO = ∠BAO by concyclicity
+      have BKO_eq_BAO : (ANG e.B e.K e.O (ne_of_not_collinear e.BKO).2.2.symm  (ne_of_not_collinear e.BKO).1).dvalue = (ANG e.B e.A e.O (ne_of_not_collinear e.BAO).2.2.symm  (ne_of_not_collinear e.BAO).1).dvalue := by
+        exact iangle_theorem e.C₅ e.BKO e.BAO e.BKOA_on_C₅
+      -- Prove ∠LKO = ∠LKB + ∠BKO
+      have LKO_eq_LKB_ad_BKO : (ANG e.L e.K e.O (ne_of_not_collinear e.LKO).2.2.symm  (ne_of_not_collinear e.LKO).1).dvalue = (ANG e.L e.K e.B (ne_of_not_collinear e.LKB).2.2.symm  (ne_of_not_collinear e.LKB).1).dvalue + (ANG e.B e.K e.O (ne_of_not_collinear e.BKO).2.2.symm  (ne_of_not_collinear e.BKO).1).dvalue := by
+        exact dvalue_add_theorem e.LKO e.LKB e.BKO
+      rw[LKB_eq_LFI.symm,BKO_eq_BAO.symm,LKO_eq_LKB_ad_BKO]
+    have LKO_eq_LNO : (ANG e.L e.K e.O (ne_of_not_collinear e.LKO).2.2.symm  (ne_of_not_collinear e.LKO).1).dvalue = (ANG e.L e.N e.O (ne_of_not_collinear e.LNO).2.2.symm  (ne_of_not_collinear e.LNO).1).dvalue := by
+      -- Prove ∠LNO = ∠LNI + ∠INO
+      have LNO_eq_LNI_ad_INO : aLNO.dvalue = aLNI.dvalue + aINO.dvalue := by
+        exact dvalue_add_theorem e.LNO e.LNI e.INO
+      -- Prove ∠LNI = ∠LFI
+      have LNI_eq_LFI : aLNI.dvalue = aLFI.dvalue := by
+        apply iangle_theorem (mk_pt_pt_pt_circle e.I e.F e.L e.IFL) e.LNI e.LFI _
+        constructor
+        exact (pt_on_mk_pt_pt_pt_circle e.IFL).2.2
+        constructor
+        exact IFNL_c
+        constructor
+        exact (pt_on_mk_pt_pt_pt_circle e.IFL).1
+        exact (pt_on_mk_pt_pt_pt_circle e.IFL).2.1
+      -- Prove ∠INO = ∠BAO
+      have INO_eq_BAO : aINO.dvalue = aBAO.dvalue := by
+        rw [iangle_theorem e.C₄ e.INO e.IAO e.INOA_on_C₄]
+        exact dvalue_eq_dir1_one_line e.l₁ e.IBA_on_l₁ e.IAO e.BAO
+      rw[LKO_eq_LFI_ad_BAO,LNI_eq_LFI.symm,INO_eq_BAO.symm,LNO_eq_LNI_ad_INO]
+    exact four_pt_circle (e.LKO) (ne_of_not_collinear e.LNO).2.2.symm (ne_of_not_collinear e.LNO).1 LKO_eq_LNO
+  -- In the same way, we can prove L,K,O,M are concyclicity
+  have LKOM_c : e.M LiesOn mk_pt_pt_pt_circle e.L e.K e.O e.LKO := by sorry
+  -- Give a conclusion
+  use mk_pt_pt_pt_circle e.L e.K e.O e.LKO
+  constructor
+  exact (pt_on_mk_pt_pt_pt_circle e.LKO).2.1
+  constructor
+  exact (pt_on_mk_pt_pt_pt_circle e.LKO).1
+  constructor
+  exact LKOM_c
+  constructor
+  exact KLON_c
+  exact (pt_on_mk_pt_pt_pt_circle e.LKO).2.2
diff --git a/project_q1/JZM_problem.png b/project_q1/JZM_problem.png
new file mode 100644
index 0000000..5f4bca4
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