EMAT 8990


Doug Westmoreland

Cyclic Quadrilaterals

(Figure 1)

Given quadrilateral ABCD.  (figure 1 above)

Consider the following:  The four angle-bisectors of a quadrilateral always determine a cyclic quadrilateral.

Quick check:  a = 180 - ( x + t )  and b = 180 - ( y + z )

                      therefore   a + b = 360 - ( x + y + z + t )

                                                 =  360 - 180

                                                 = 180

So,  quadrilateral EFGH is a cyclic quadrilateral.

Click here for a GSP sketch of a cyclic quadrilateral that you can drag around and investigate.

Try dragging the vertices of quadrilateral ABCD so that the quadrilateral is a parallelogram.  What do you think about the cyclic quadrilateral that is formed?  If it is a special type of quadrilateral, can you prove the special case?

Further investigations on the cyclic quadrilateral

Consider this:

The four lines, each drawn from the midpoint of a side of a cyclic quadrilateral perpendicular to the opposite side, are concurrent.     Can you prove this?  Are there any special cases to consider?

Click here for a GSP model to investigate.


Consider this:

Brahmagupta's Theorem:  In a cyclic quadrilateral having perpendicular diagonals, the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

Note: Brahmagupta was a Hindu mathematician 628 A.D. The position of the anticenter, the point of intersection of the diagonals of the cyclic quadrilateral, in the particular case of a cyclic quadrilateral with perpendicular diagonals was the discovery of Brahmagupta.

Click here for a GSP model to investigate.

Can you prove Brahmagupta's Theorem?  Click here for a hint.

Investigations on your own

Use GSP to make sketches of the following situations and see if you can prove them.

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