We begin our proof by drawing a rectangle ABCD. We then draw CE outside the rectangle so that AD congruent CE. Point P is the intersectoin of the perpendicular bisectors of AE and CD, which intersect AE and CD at points M and N, respectively. Drawing DP, AP, EP, and CP completes the diagram for this "proof".
Because AP = EP and DP = CP (every point on theperpendicular bisector of a line segment is equidistant from the endpoints of theline segment), we have triangle ECP is congruent to triangle ADP (SSS) and the measure of angle ECP = the measure of angle ADP.
However, because triangle PDC is isosceles, the measure of angle DCP = the measure of angle CDP. By subtraction, obtuse angle ECD has the same measure as angle ADC, a right angle!!
You may wish to consider the case when P is on DC or when P is in rectangle ABCD. Similar arguments hold for these cases.
By now you may find that an accurate construction is the best way to isolate the error in the 'proof'. Rahter than attempt to discover the error by construction, we will analyze the situation that now exists. We notice that NP is also the perpendicular bisector of AB. Consider triangle ABE. Because NP and MP are the perpendicular bisectors of AB and AE, respectively, they intersect at the center, P, of the circumcircle of triangle ABE. Therefore, point P must also be on the perpendicular bisector of BE.
By construction, we have BC = EC. Therefore point C must also lie on the perpendicular bisector of BE.
PC is the perpendicular bisector of BE as well as the interior angle bisector of angle BCE. A reflex angle is an angle of measure greater than 1880 degrees and less than 360 degrees. Consider reflex angle ECP, whose measure is measure of angle PCR + measrue of angle RCE. Thus EP in triangle ECP is placed so that it is on the side of point C outside the rectangle. This makes the last step of our proof incorrect because measure of angle ECP does not equal the measure of angle ECD + the measure of angle DCP.
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