 EMAT 8990

by

Doug Westmoreland (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

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?

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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.

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

• In a cyclic quadrilateral with perpendicular diagonals, the distance from the circumcenter to a side is just half the opposite side.

• Let the diagonals of cyclic quadrilateral ABCD meet at M and let their midpoints be E and F. Then the orthocenter of triangle EMF is the anticenter T of ABCD.

• Let P and Q be the midpoints of a pair of opposite sides, say AB and CD.  The centroid K of ABCD is not only the midpoint of PQ but is also the midpoint of the segment EF joining the midpoints of the diagonals.