Forming an Arbitrary Quadrilateral

[BMAD]

The four squares connect at corners defining an arbitrary quadrilateral. Connecting the centers of opposite squares, show that the segments are perpendicular.

[BMAD]

Represent the vertices of the quadrilateral by complex numbers. Find expressions for the two vectors between the centers of opposite squares.

[BMAD]

vector algebra is also a popular method to solving this problem

[TimeSpaceLightForce]

we'll check the right angleness of B , so with A (measured)

a + A =180deg , c from right triangle, d from right triangle.

for parallelogram inside angles: 360 - a - c - d = B = 90deg..

[TimeSpaceLightForce]

..if we align the yellow square side with the blue square side..

angles d' & c' ( other angles of the right triangles) will become 0 deg

thus 90 -A = c' + d' = (90-c) + (90-d)=180 - c - d

or c + d = 90 + A...

[TimeSpaceLightForce]

previously..
eq1: 360 - (180-A) - c - d = B
eq2: c + d = 90 + A

if A2 , A3 & A4 are the other outside angles of squares
while e & f are the two other angles of right triangles.
equating counterclockwise

eq3: 360 -(180-A2) - (180 - c) - e = B (B,B2,B3,B4:same)
eq4: (180-c) + e = 90 + A2
eq5: 360 - (180-A3) - (180-e) - f = B
eq6: (180-e) + f = 90 +A3
eq7: 360 - (180-A4) - (180-f) - (180-d) =B
eq8: (180-f) + (180 - d) = 90 + A4

we have 9 variables but just 8 equations.. so finding
relations of As..if the squares are all of the same sizes,
the inside angles is 360 = outside angles , therefore
eq9: 360 = A + A2 + A3 +A4

...solving B

Edited May 12, 2013 by TimeSpaceLightForce

[TimeSpaceLightForce]

simplifying ...eq1,eq3,eq5,eq7
180+A-c-d=B
A2+c-e=B
A3+e-f = B
A4+f+d-180 = B

adding the left & right expressions
180 + A - c - d + A2 + c - e + A3 + e - f + A4 + f + d -180 = 4B
A+ A2+A3+ A4 = 4B = 360
B=90 !

[BMAD]

Represent the vertices of the quadrilateral by the complex numbers 2a, 2b, 2c, 2d, in the complex plane.

Then m, the midpoint of the line connecting 2a and 2b, is represented by a + b.
The vector from 2a to m is given by (a + b) − 2a = b − a.
Rotating this vector clockwise through 90° yields mw, the vector from m to the center of the square, w.
A rotation through 90° clockwise is achieved by multiplying by −i.
Hence the vector mw is given by −i(b − a) = (a − b)i.
Therefore w = a + b + (a − b)i.

We can similarly derive the complex number that represents the center of the other squares, yielding:

w = a + b + (a − b)i
x = b + c + (b − c)i
y = c + d + (c − d)i
z = d + a + (d − a)i

Hence the line segments joining the centers of opposite squares are given by the following two vectors:

y − w = c + d − a − b + (c − d − a + b)i z − x = d + a − b − c + (d − a − b + c)i

Therefore z − x = −i(y − w).
This tells us that vector xz is obtained by rotating wy through 90°.
Therefore line segments wy and xz lie on perpendicular lines and are of equal length.