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In the diagram below, M is the midpoint of AB. Find MN


 Oct 2, 2019

Does this problem require trigonometry? This is very hard problem!            This problem should use similar triangles i believe




MN = x with triangle AMN similar to triangle ABX


Triangle ANC is proportional to AXC?


So set up proportion




With y being length of NX



How do we go from here????






If M is the mid point, then can't we just do \(\frac{x}{14}=\frac{1}{2}\)


x = 7           So MN = 7??????

 Oct 3, 2019
edited by CalculatorUser  Oct 3, 2019
edited by CalculatorUser  Oct 3, 2019
edited by CalculatorUser  Oct 3, 2019

Maybe not as difficult as it seems.....


Note that angles XNC  and ANC  are equal

And angles ACN and XCN  are equal

And NC = NC


Therefore, by ASA, triangles ANC and XNC are congruent


Then  AN  = XN  =  5     ....so.... AX  = 10


So ....N is the midpoint of  AX


Then.......if we connect MN.......this segment is paralell  to  BX because  MN splits the sides of triangle  ABX proportionally....that is   AM/BM  = AN/XN 


So  because MN is parallel to BX, then  triangle  AMN  is similar to triangle ABX  which implies that


MN / BX  =  AN / AX


MN  / 14   =  5 / 10


MN /14  = 1/2


MN  = (1/2) (14)    = 7   as found by CU!!!


cool cool cool

 Oct 3, 2019
edited by CPhill  Oct 3, 2019

XD my answer was luck!!

CalculatorUser  Oct 4, 2019
edited by CalculatorUser  Oct 6, 2019

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