x-axis is the trapezoid midline parallel to the bases and y-axis is common perpendicular bisector of the bases. $ U, V$ are midpoints of $ DA, BC.$ Let $ DA = 2u, BC = 2v, UV = 2h.$ At time $ t = 0,$ points $ P, Q \in AB$ pass through $ A, B$ in the opposite directions with constant velocities proportional to $ u, v$; let their x-coordinates be $ x_P = u(1 - t)$ and $ x_Q = v(1 + t).$ Equation of the line $ AB$ yields their y-coordinates, and the y-coordinate of the midpoint $ R$ of $ PQ$: $ y + h = - \frac {2h}{u - v}(x - u)$ or $ y - h = - \frac {2h}{u - v} (x - v),$ $ y_P = \frac {2hut}{u - v} - h$ and $ y_Q = h - \frac {2hvt}{u - v}$ $ \Longrightarrow$ $ y_R = ht.$ Equations of the lines $ DP, CQ$ are then $ y + h = \frac {2ht}{(u - v)(2 - t)} (x + u)$ and $ y - h = - \frac {2ht}{(u - v)(2 + t)}(x + v).$ Multiplying by $ (2 - t), (2 + t),$ respectively, and adding the 2 equations eliminates $ x,$ yielding the y-coordinate of their intersection $ S$: $ (y + h)(2 - t) + (y - h)(2 + t) = 2ht\ \Longrightarrow\ y_S = ht$ Since $ y_S = y_R,$ it follows that $ RS \parallel DA \parallel BC.$ Since $ \frac {AM}{BM} = \frac {DA}{BC} = \frac {u}{v},$ the points $ P, Q, R, S$ coincide with $ X, Y, Z, N,$ respectively, at the same time and $ ZN \parallel DA \parallel BC.$

EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ X,\ Y$ be, the intersection points of its sidelines $ AB,\ AC,$ from the lines through $ C,\ B$ respectively and parallel to the line segment $ AN,$ as the internal angle bisector of $ \angle A.$ We denote the points $ D,\ E$ on the side-segment $ BC$ such that $ CD = AC,\ BE = AB$ and let be the point $ Q\equiv DX\cap EY.$ Prove that the line through $ Q$ and parallel to $ AN,$ bisects the segment $ DE.$ GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on the line segment $ AN,$ as the internal angle bisector of $ \angle A$ $ ($ $ A,$ between $ P$ and $ N\in BC$ $ ).$ We denote the points $ X\equiv AB\cap PC,\ Y\equiv AC\cap PB$ and let $ D,\ E$ be, two points on the side-segment $ BC$ such that $ CD = AC,\ BE = AB.$ Prove that the line segment $ PQ,$ where $ Q\equiv DX\cap EY,$ bisects the segment $ DE.$ I will post here later, the solution of the general problem I have in mind. Kostas Vittas.

Attachments:
t=215967(a).pdf (5kb)
t=215967.pdf (4kb)

Can you please do so, vittasko?

vittasko wrote: EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ X,\ Y$ be, the intersection points of its sidelines $ AB,\ AC,$ from the lines through $ C,\ B$ respectively and parallel to the line segment $ AN,$ as the internal angle bisector of $ \angle A.$ We denote the points $ D,\ E$ on the side-segment $ BC$ such that $ CD = AC,\ BE = AB$ and let be the point $ Q\equiv DX\cap EY.$ Prove that the line through $ Q$ and parallel to $ AN,$ bisects the segment $ DE.$ PROOF. – (see the schema t=215967) From $ QM\parallel CX$ $ \Longrightarrow$ $ \frac{MD}{CD} = \frac{QM}{CX}$ $ ,(1)$ From $ QM\parallel BY$ $ \Longrightarrow$ $ \frac{ME}{BE} = \frac{QM}{BY}$ $ ,(2)$ From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac{MD}{ME}\cdot \frac{BE}{CD} = \frac{BY}{CX}$ $ \Longrightarrow$ $ \frac{MD}{ME} = \frac{AC}{AB}\cdot \frac{BY}{CX}$ $ ,(3)$ But, from $ BY\parallel CX$ $ \Longrightarrow$ $ \frac{AC}{AB} = \frac{CX}{BY}$ $ ,(4)$ Hence, from $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac{MD}{ME} = 1$ $ \Longrightarrow$ $ MD = ME$ and the proof is completed. vittasko wrote: GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on the line segment $ AN,$ as the internal angle bisector of $ \angle A$ $ ($ $ A,$ between $ P$ and $ N\in BC$ $ ).$ We denote the points $ X\equiv AB\cap PC,\ Y\equiv AC\cap PB$ and let $ D,\ E$ be, two points on the side-segment $ BC$ such that $ CD = AC,\ BE = AB.$ Prove that the line segment $ PQ,$ where $ Q\equiv DX\cap EY,$ bisects the segment $ DE.$ PROOF. - Let be the point $ Z\equiv BC\cap XY$ and it is easy to show that this point is the harmonic conjugate of $ N,$ with respect to $ B,\ C$ $ ($ from the complete quadrilateral $ PXAYBC$ $ )$ and so, we conclude that the line segment $ AZ,$ is the external angle bisector of $ \angle A$ of the given triangle $ \bigtriangleup ABC.$ So, we have $ \frac {NB}{NC} = \frac {ZB}{ZC} = \frac {AB}{AC}$ $ ,(1)$ $ \bullet$ We see that the pencils $ X.PYQM,\ Y.PXQM,$ have the line segment $ XY,$ as their common ray and the points $ P\equiv XP\cap YP,\ Q\equiv XQ\cap YQ,\ M\equiv XM\cap YM$ are collinear and so, they have congruent double ratios ( = cross ratios ). Hence, we conclude that $ (X.PYQM) = (Y.PXQM)$ $ ,(2)$ The pencils $ X.PYQM,\ Y.PXQM,$ are intersected from the sideline $ BC$ of $ \bigtriangleup ABC$ and so, we have: $ (X.PYQM) = (C,Z,D,M)$ $ ,(3)$ and $ (Y.PXQM) = (B,Z,E,M)$ $ ,(4)$ From $ (2),$ $ (3),$ $ (4)$ $ \Longrightarrow$ $ (C,Z,D,M) = (B,Z,E,M)$ $ ,(5)$ From $ (5)$ $ \Longrightarrow$ $ \frac {DC}{DZ} \div \frac {MC}{MZ} = \frac {EB}{EZ} \div \frac {MB}{MZ}$ $ \Longrightarrow$ $ \frac {AC}{DZ} \cdot \frac {1}{MC} = \frac {AB}{EZ} \cdot \frac {1}{MB}$ $ \Longrightarrow$ $ \frac {AB}{AC} = \frac {MB}{MC}\cdot \frac {ZE}{ZD}$ $ ,(6)$ From $ (1),$ $ (6)$ $ \Longrightarrow$ $ \frac {ZB}{ZC} = \frac {MB}{MC}\cdot \frac {ZE}{ZD}$ $ \Longrightarrow$ $ \frac {ZB}{ZC} = \frac {MB}{MC}\cdot \frac {ZB + AB}{ZC - AC}$ $ ,(7)$ From $ \frac {ZB}{ZC} = \frac {AB}{AC},$ it is easy to show that $ \frac {ZB - AB}{ZC - AC} = \frac {AB}{AC} = \frac {ZB}{ZC}$ $ ,(8)$ From $ (7),$ $ (8)$ $ \Longrightarrow$ $ \frac {ZB - AB}{ZC - AC} = \frac {MB}{MC}\cdot \frac {ZB + AB}{ZC - AC}$ $ \Longrightarrow$ $ \boxed{\frac {MB}{MC} = \frac {ZB - AB}{ZB + AB}}$ $ ,(9)$ $ \bullet$ Let $ M'$ be, the midpoint of the segment $ DE$ and we will prove that $ M'\equiv M.$ We put $ BC = a,\ AC = b,\ AB = c$ and then we have: $ BD = a - b$ $ CE = a - c$ $ DE = a - BD - CE = b + c - a$ It is easy to show now, that $ M'B = BD + \frac {DE}{2} = \frac {a - b + c}{2}$ $ ,(10)$ and $ M'C = CE + \frac {DE}{2} = \frac {a + b - c}{2}$ $ ,(11)$ From $ (10),$ $ (11)$ $ \Longrightarrow$ $ \boxed{\frac {M'B}{M'C} = \frac {a - b + c}{a + b - c}}$ $ ,(12)$ From $ (9),$ $ (12),$ it is enough to prove that: $ \frac {ZB - AB}{ZB + AB} = \frac {a - b + c}{a + b - c}$ $ \Longrightarrow$ $ \frac {ZB - c}{ZB + c} = \frac {a - b + c}{a + b - c}$ $ \Longrightarrow$ $ \boxed{ZB = \frac {ac}{b - c}}$ $ ,(13)$ But the relation $ (13)$ is true from $ \frac {ZB}{ZC} = \frac {AB}{AC} = \frac {c}{b},$ where $ ZC = ZB + a$ So, we conclude that $ M'\equiv M$ and hence, the point $ M\equiv BC\cap PQ$ is the midpoint of the segment $ DE$ and the proof is completed. Kostas Vittas.

Attachments:
t=215967(b).pdf (5kb)

Proof of initial problem: Let {P}=AD∩CY and Q = BC∩DX. Obviously, AD/BC=AM/BM ( 1 ), since CM=BM. We have also AP/BC=AY/BY=(AB-BM)/BM ( 2 ). From (1) and (2) we get PD/BC=(AB-BM+AM)/BM ( 3 ); similarly CQ/AD=(AB-AM+BM)/AM ( 4 ). From (3) and (4) we get PN/CN=PD/CQ=(AB-BM+AM)/(AB+BM-AM) ( 5 ). By straight calculations we get AZ/BZ= (AB-BM+AM)/(AB+BM-AM) ( 6 ), so PN/CN=AZ/BZ , q.e.d. Proof of general problem: Let {M}=DX∩PE and {Q}=EY∩PD. Applying Menelaus in triangle PNC with the transversal BAX we get PX/CX=c.AP/(b+c).AN ( 1 ), after having taken into account that AN was the angle bisector of ÐA; similarly PY/CY=b.AP/(b+c).AN ( 2 ) (a, b and c being the sides of triangle ABC). Next, apply Menelaus to triangle PCE with the transversal DMX and get bc.AP/(b+c)(b+c_a).AN ( 3 ); similarly, apply Menelaus to triangle PBD with the transversal EQY and get for PQ/QE the same value ( 4 ). From (4) and (3) we get that DM and EQ intersect on the median from P of the triangle PDE. Best regards, sunken rock