Let ABCD be an isosceles trapezoid with AD∥BC. Its diagonals AC and BD intersect at point M. Points X and Y on the segment AB are such that AX=AM, BY=BM. Let Z be the midpoint of XY and N is the point of intersection of the segments XD and YC. Prove that the line ZN is parallel to the bases of the trapezoid. Author: A. Akopyan, A. Myakishev
Problem
Source: Tuymaada 2008, Junior League, Second Day, Problem 6.
Tags: geometry, trapezoid, analytic geometry, ratio, perpendicular bisector, angle bisector, projective geometry
21.07.2008 09:26
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∈AB pass through A,B in the opposite directions with constant velocities proportional to u,v; let their x-coordinates be xP=u(1−t) and xQ=v(1+t). Equation of the line AB yields their y-coordinates, and the y-coordinate of the midpoint R of PQ: y+h=−2hu−v(x−u) or y−h=−2hu−v(x−v), yP=2hutu−v−h and yQ=h−2hvtu−v ⟹ yR=ht. Equations of the lines DP,CQ are then y+h=2ht(u−v)(2−t)(x+u) and y−h=−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 ⟹ yS=ht Since yS=yR, it follows that RS∥DA∥BC. Since AMBM=DABC=uv, the points P,Q,R,S coincide with X,Y,Z,N, respectively, at the same time and ZN∥DA∥BC.
22.07.2008 00:18
EQUIVALENT PROBLEM. – A triangle △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 ∠A. We denote the points D, E on the side-segment BC such that CD=AC, BE=AB and let be the point Q≡DX∩EY. Prove that the line through Q and parallel to AN, bisects the segment DE. GENERAL PROBLEM. – A triangle △ABC is given and let P be, an arbitrary point on the line segment AN, as the internal angle bisector of ∠A ( A, between P and N∈BC ). We denote the points X≡AB∩PC, Y≡AC∩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≡DX∩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)
17.08.2008 18:19
Can you please do so, vittasko?
26.08.2008 14:05
vittasko wrote: EQUIVALENT PROBLEM. – A triangle △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 ∠A. We denote the points D, E on the side-segment BC such that CD=AC, BE=AB and let be the point Q≡DX∩EY. Prove that the line through Q and parallel to AN, bisects the segment DE. PROOF. – (see the schema t=215967) From QM∥CX ⟹ MDCD=QMCX ,(1) From QM∥BY ⟹ MEBE=QMBY ,(2) From (1), (2) ⟹ MDME⋅BECD=BYCX ⟹ MDME=ACAB⋅BYCX ,(3) But, from BY∥CX ⟹ ACAB=CXBY ,(4) Hence, from (3), (4) ⟹ MDME=1 ⟹ MD=ME and the proof is completed. vittasko wrote: GENERAL PROBLEM. – A triangle △ABC is given and let P be, an arbitrary point on the line segment AN, as the internal angle bisector of ∠A ( A, between P and N∈BC ). We denote the points X≡AB∩PC, Y≡AC∩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≡DX∩EY, bisects the segment DE. PROOF. - Let be the point Z≡BC∩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 ∠A of the given triangle △ABC. So, we have NBNC=ZBZC=ABAC ,(1) ∙ We see that the pencils X.PYQM, Y.PXQM, have the line segment XY, as their common ray and the points P≡XP∩YP, Q≡XQ∩YQ, M≡XM∩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 △ABC and so, we have: (X.PYQM)=(C,Z,D,M) ,(3) and (Y.PXQM)=(B,Z,E,M) ,(4) From (2), (3), (4) ⟹ (C,Z,D,M)=(B,Z,E,M) ,(5) From (5) ⟹ DCDZ÷MCMZ=EBEZ÷MBMZ ⟹ ACDZ⋅1MC=ABEZ⋅1MB ⟹ ABAC=MBMC⋅ZEZD ,(6) From (1), (6) ⟹ ZBZC=MBMC⋅ZEZD ⟹ ZBZC=MBMC⋅ZB+ABZC−AC ,(7) From ZBZC=ABAC, it is easy to show that ZB−ABZC−AC=ABAC=ZBZC ,(8) From (7), (8) ⟹ ZB−ABZC−AC=MBMC⋅ZB+ABZC−AC ⟹ MBMC=ZB−ABZB+AB ,(9) ∙ Let M′ be, the midpoint of the segment DE and we will prove that M′≡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+DE2=a−b+c2 ,(10) and M′C=CE+DE2=a+b−c2 ,(11) From (10), (11) ⟹ M′BM′C=a−b+ca+b−c ,(12) From (9), (12), it is enough to prove that: ZB−ABZB+AB=a−b+ca+b−c ⟹ ZB−cZB+c=a−b+ca+b−c ⟹ ZB=acb−c ,(13) But the relation (13) is true from ZBZC=ABAC=cb, where ZC=ZB+a So, we conclude that M′≡M and hence, the point M≡BC∩PQ is the midpoint of the segment DE and the proof is completed. Kostas Vittas.
Attachments:
t=215967(b).pdf (5kb)
20.03.2009 15:23
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