Thread: Paralellogram

Any cyclic paralellogram having unequal adjacent sides is necessarily a
(a)Square
(b)Rectangle
(c)Rhombus
(d)Trapezium
Why?

If a parallelogram is cyclic, then it must be a rectangle. The perpendicular bisectors of the sides of a parallelogram are concurrent only when it is a rectangle. For a parallelogram to be cyclic, its perpendicular bisectors must be concurrent at the center of the circumcircle; therefore, a cyclic parallelogram must be a rectangle.

Hello, devi!

Any cyclic paralellogram having unequal adjacent sides is necessarily a:
. . (a) square . (b) rectangle . (c) rhombus . (d) trapezium

This is my approach to this problem.
Perhaps you can hammer it into a formal proof.


Opposite sides of a parallelogram are parallel.
They are formed by a pair of parallel chords.

Code:

              * * *
          *           *
        *               *
     A o o o o o o o o o o B
                  
      *                   *
      *         *         *
    D o o o o o o o o o o o C
                  
       *                 *
        *               *
          *           *
              * * *

Opposite sides of a parallelogram are equal.
They are formed by two chords equidistant from the center.

Code:

              * * *
          *           *
      A o o o o o o o o o B
       *        :        *
                :
      *         :         * C
      *         *         o
      *          \      o *
                  \   o
       *            o    *
        *         o     *
          *     o     *
              o * *
            D

Since opposite sides of a parallelogram are parallel and equal,
the chords are parallel and equidistant from the center.

Code:

              * * *
          *           *
      A o o o o o o o o o B
       *        :        *
                :
      *         :         *
      *         *         *
      *         :         *
                :
       *        :        *
      D o o o o o o o o o C
          *           *
              * * *

Your turn . . .