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 1182.

## Thursday, January 28, 2016

### Geometry Problem 1182: Two Squares, Four Quadrilaterals, Sum of the Areas, Metric Relations

Labels:
area,
metric relations,
quadrilateral,
square

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draw from E parallel to AD and to AB, at the same way from F, G, H

ReplyDeleteSideways EF, FG,GH,HE are formed 4 congr triangles, so and sideways

of a, b, c, d

And results is clear

http://s21.postimg.org/nszfowro7/pro_1182.png

ReplyDeleteDraw new square E’F’G’H’ as shown on the sketch

Let x= FF’=GG’=HH’=EE’

Define u, v, s, t as shown

Denote S(XYZ)= area of XYZ

Let S5= S(GFF’)=S(FEE’)=S(EHH’)=S(HGG’)

1. Note that u+v=s+t

With algebra calculation we get

.a^2 +c^2= u^2+v^2+s^2+t^2+2.x^2+2x(s+t)

And b^2+d^2= u^2+v^2+s^2+t^2+2.x^2+2x(u+v)

Since u+v= s+t => a^2 +c^2= b^2+d^2

2. Let S1’=S(BF’E’A), S2’=S(BF’G’C), S3’=S(CG’H’D) , S4’=S(AE’H’D)

Since u+v=s+t => S1’+S3’=S2’+S4’…. (1)

We have S1=S1’+S5+1/2x(t-s)

And S3=S3’+S5+1/2x(s-t)

So S1+S3=S1’+S3’+2.S5…. (2)

Similarly S2+S4=S2’+S4’+2.S5…(3)

Compare (1),(2) and (3) we will get S1+S3=S2+S4