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fiora

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fiora  Mar 20, 2019
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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 angle MDA + angle BEA = 90 degrees  angle AND and angle MNA are right angle

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

MA / AD =1/2   ND = 2*AN and AN = 2* MN  DN / MN =4

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

BF=52a  by pythagorean theorem, cos of angle CBF = BC / BF = 25, BG = BE *cos CBF =1/2 * a*2/sqrt(5)=55a 

GF= BF - GF =3510a, GE=1/2*BG=510a,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 = 25a

Triangle BAG is similar MAN. AM/AB=1/2  AN= 1/2 AG  GN= 1/2*AG = 55a

BN= (55a)2+(55a)2=105a by  pythagorean theorem, and MN =EG=510a

BN/MN22

 

Edit: highlighted answer

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Feb 24, 2020
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Oct 25, 2018