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

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

Saketh Oct 2, 2019

#1**+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??????

CalculatorUser Oct 3, 2019

edited by
CalculatorUser
Oct 3, 2019

edited by CalculatorUser Oct 3, 2019

edited by CalculatorUser Oct 3, 2019

edited by CalculatorUser Oct 3, 2019

edited by CalculatorUser Oct 3, 2019

#2**+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!!!

CPhill Oct 3, 2019