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

https://latex.artofproblemsolving.com/2/4/e/24ee65782c8200565297afc86faac0de3368ec25.png

 Oct 2, 2019
 #1
avatar+2417 
+1

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

 

\(\frac{x}{14}=\frac{5}{5+y}\)

 

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
 #2
avatar+104962 
+2

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
 #3
avatar+2417 
+2

XD my answer was luck!!

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

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