What Can You Prove? (Posted on 2004-01-14)

• In the figure, above AC=BD.
  
  
• ‹ACD is a right angle.
  
  
• ‹BDC is less than 90º.
  
  
• QP and RP are perpendicular bisectors of AB and CD respectively, intersecting at P.
  
  
• AP=BP and CP=DP because all points on the perpendicular bisector of a line are equidistant from the extremities of the line.
  
  
• Triangle APC is therefore congruent to triangle BPD because of the well known "side-side-side" theorem. Now, ‹PCD=‹PDC because the base angles of an isosceles triangle are equal.
  
  
• ‹ACP=‹BDP because they are corresponding parts of congruent triangles.
  
  
• Of course, ‹ACP + ‹PCD = ‹BDP + ‹PDC because when equals are added to equals the results are equal.

Therefore, ‹ACD = ‹BDC.

But, wait, we know that this isn't true!

A mathematician had a rickety table (with top AB sitting on the floor CD with legs AC and BD) and had constructed the drawing above in an ineffectual effort to straighten the table. Can you figure out what is wrong?

	Submitted by DJ
Rating: 4.1818 (11 votes)
Solution:	(Hide)
When you actually draw PQ as the perpendicular bisector of AB, point P is relatively high, depending on just how acute angle BDC is. Whatever the angle, this point is always high enough so that, above the angle of BDC, segment PD is actually external to ABDC, and BDC is actually the difference between PCD (=PDC) and that angle (PDB). See the diagram when the "acuteness" of ‹BDC is exaggerated:

	Subject	Author	Date
	to SK	luminita	2004-01-23 17:08:49
	Why do you need Z?	Dan Porter	2004-01-18 06:10:14
	re: solution	SilverKnight	2004-01-16 17:12:45
	solution	luminita	2004-01-16 16:36:09
	solution	Dan Porter	2004-01-15 06:53:51
	No Subject	Dan Porter	2004-01-15 06:53:16
	re: solution	Sam	2004-01-14 23:43:37
	solution	Charlie	2004-01-14 09:15:28