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