Two Rectangles in a Rectangle
In the applet, ABCD, EFGH, and IFJH are all rectangles. Show that
 |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny
Since the diagonals of a rectangle (and even those of a parallelogram) meet at the common midpoint and since the two internal rectangles share one diagonal FH, their diagonals are equal and centers coincide. This point O is also the center of ABCD. Indeed, being the midpoint of FH it lies on the horizontal axis of ABCD. Being the midpoint of, say, EG, it lies on the vertical axis of ABCD. We found that EG = IG and both pass through the center of ABCD. Hence, triangle GOJ is isosceles and
such that BCGI is another rectangle and AIGD ie yet another one. Obviously, Area(IFG) = Area(BCGI)/2 and Area(IGH) = Area(AIGD)/2. It follows that the area of the quadrilateral IFGH is half that of ABCD. The same of course is true for the quadrilateral EFJH:
We are in the framework of a variant of the Carpets Theorem. With The second result is an immediate consequence of the first. References
Carpets Theorem
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
| 40619608 |
