Consider trapezoid ABCD where BC is perpendicular to AB and CD. Consider also that the diagonals AC and BD are perpendicular to each other and intersect at point E.
If EA = 6.25 and ED = 3.2, then what are EB and EC?
I did not notice the fact that triangle CEB also participates in the similarity between DEC and EBA. Instead ...
Call mCD = b; mCB = a; mBA = c
mCE = x; mEB = y
By the similarity of DEC with BEA, b/c = 3.2/y
Also since 3.2/y=x/6.25, xy = 20.
The upward slope of one diagonal is a/b, and of the other, a/c, so, since they are perpendicular, b/a = a/c, or bc = a^2
Also by Pythagoras on DEC, 3.2^2+x^2 = b^2.
Setting up cells in Excel for a, b, c, x and y, and others with formulae for b/c, 3.2/y, x*y, a^2, x^2+y^2 and b^2, and then using Solver to set a, b, c, x and y to make the xy cell equal to 20, while fitting in the above constraints, finds CE=x = 4 and EB=y = 5 along with CB=a=6.40312423743285, CD=b=5.12249938994628, BA=c=8.00390529679105.
This agrees with the results from playing around with the figure in Geometer's Sketchpad.
Edited on March 3, 2007, 12:02 pm

Posted by Charlie
on 20070303 12:00:45 