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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1185.
Draw new rectangle E’F’G’H’ similar to problem 1182Let k= AD/AB=EH/EFLet x= FF’=HH’ =>EE’=GG’=k.xAll notations u, v,s, t , S(XYZ) will be the same as problem 1182Perform length and width comparison of 2 similar rectangles ABCD and EFGH we will get(u+v+x)/(t+s+k.x)= k …… (1)Similar to pro. 1182 we have a^2+c^2= u^2+v^2+s^2+t^2+2k.x(k.x+t+s)…..(2)And b^+d^2= u^2+v^2+s^2+t^2+2x(x+u+v)…..(3)Replace expression (1) in (2) we will get a^2+c^2=b^2+d^2
http://s25.postimg.org/5ag8xy18f/pro_1185_2.pngSee below for solution of problem 1185 part 2From midpoint of each side of rectangle EFGH draw a new rectangle (red) with each side parallel to rectangle ABCD.Define u’, v’, s’, t’ , z, w as shown on the sketchNote that this red rectangle is similar to rect. ABCD and(u’+v’)/(s’+t’)= AD/AB= z/w ….. (4)At each corner of ABCD we have 2 equal areas triangles so the problem become to show that trapezoids areas of (S1+S3)= trapezoids areas of ( S2+S4). Trapezoids areas of (S1+S3)= w(u’+v’)And trapezoids areas of ( S2+S4).= z(s’+t’).Using relation of (4) in above expressions we will get S1+S3=S2+S4
Second half of solution draw EE' perpendicular to AB and EE`` to AD.At the same way from F,G,HTrapezoids EFF'E', GHH``G`` have altitude E'F' trapezoids FGF``G`` and EHH``E``have altitude E``H``. Ratio E'F'/E``H``= AB/AD = EF/FH (similar tr with perp sides)Sum of their bases are in ratio AD/AB=FH/EF (as diferences of inverse ratio)First part of solutions is clear acording to pythaghor theorem included ratiosabove
Sum of S of trapezoids (' on AB,CD, " on AD, BC)E"H"(AB-E'F'+AB-E'F')=? E'F'(AD-E"H"+AD-E"H")E"H"AB-E"H"E'F' =? E'F'AD-E'F'E"H"E"H"AB =? E'F'ADAB/AD = E'F'/E"H"