Disguised trapezoid (Posted on 2005-05-28)

	Submitted by McWorter
Rating: 4.2000 (5 votes)
Solution:	(Hide)
If you know about cyclic quadrilaterals (if you don't, a different proof follows), ABCD is cyclic because opposite angles are supplementary. Thus all vertices of ABCD lie on a circle. Since AB and CD are equal chords on that circle, angles ACB and CAD are equal. Hence AD and BC are parallel. A proof without knowledge of cyclic quadrilaterals might go like this. If AD and BC have the same length, then ABCD is a parallelogram and we are done. Otherwise, without loss of generality we may assume AD and BC have different lengths a and b, resp., with b<a. The AB and CD meet in a point E, with C between E and D. We show that sides AE and ED of triangle AED are of equal length. Let x be the lengths of AB and CD, y be the length of BE, and z be the length of EC. Since angles BAD and BCD are supplementary, angle BAD equals angle BCE. Hence the triangles AED and BCE are similar, having the equal angles BAD and BCE and the common angle BED. By similar triangles, noting that AE has length x+y and ED has length x+z, (x+z)/y=(x+y)/z, or xz+z^2=xy+y^2, or xz-xy=y^2-z^2, or x(z-y)=(y-z)(y+z)=(z-y)(-y-z). If y and z are not equal, then x=-y-z. But then x is negative, which is impossible; so y must equal z. Hence triangle BEC is isosceles and so angle EBC equals angle BCE which is the same as angle BAD. Hence AD is parallel to BC.

	Subject	Author	Date
	Cyclic	Bractals	2005-05-29 01:50:35
	Solution by Contradiction	Victor Zapana	2005-05-28 18:12:59
	One solution	Gamer	2005-05-28 17:09:21