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M is the midpoint of BA, F is the midpoint of DC, and E is the midpoint of CB. Find BN/MN, FG/MN, and DN/MN. I know we have to use similar triangles but I don't know exactly how to do the problem. Please explain it to me so I can understand. I know you can use coordinate geometry, but I want to know if there's another way. Thanks in advance!

 Feb 24, 2020
edited by Guest  Feb 24, 2020
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Assume the quadrilateral is a square, then AB = BC = CD = DA =a for some non-negative a

AM = BM = BE =EC = CF = FD = 1/2*a (given)

Triangle MAD is congruent to triangle EBA also congruent to FCB (Side-angle-side)

Therefore, angle EAB = angle MDA, angle AMD = angle BEA (corresponding angles are equal)

angle EAB + angle BEA = 90 degrees \(\Rightarrow\)angle MDA + angle BEA = 90 degrees \(\Rightarrow\) angle AND and angle MNA are right angle

That is triangle MNA is similar to triangle AND and triangle MAD

MA / AD =1/2  \(\Rightarrow\) ND = 2*AN and AN = 2* MN \(\Rightarrow\) DN / MN =4

Similarly, you can prove angle BGE is a right angle, which yield Trinagle BGE is similar to BCF (Angle-Angle)

BF=\(\frac{\sqrt{5}}{2}a\)  by pythagorean theorem, cos of angle CBF = BC / BF = \(\frac{2}{\sqrt{5}}\), BG = BE *cos CBF =1/2 * a*2/sqrt(5)=\(\frac{\sqrt5}{5}a\) 

GF= BF - GF =\(\frac{3\sqrt5}{10}a\), GE=1/2*BG=\(\frac{\sqrt5}{10}a\),FG/GE = 3 , Triangle BGE is congruent to triangle ANM, which yield MN =EG

Therefore FG/MN= FG/GE = 3

Connect BN, AG = AB*cos BAG = AB * cos CBF = \(\frac{2}{\sqrt{5}}a\)

Triangle BAG is similar MAN. AM/AB=1/2 \(\Rightarrow\) AN= 1/2 AG \(\Rightarrow\) GN= 1/2*AG = \(\frac{\sqrt5}{5}a\)

BN= \(\sqrt{{​​(\frac{\sqrt5}{5}a)}^{2}+{​​(\frac{\sqrt5}{5}a)}^{2}}=​​\frac{\sqrt{10}}{5}a\) by  pythagorean theorem, and MN =EG=\(\frac{\sqrt5}{10}a\)

BN/MN\(2\sqrt{2}\)

 

Edit: highlighted answer

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 Feb 25, 2020
edited by fiora  Feb 25, 2020

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