Geometrical fallacy
21 May '04 12:56How to prove that a right angle is bigger than a right angle!
This requires a pencil and ideally an A4 sheet of paper to do a construction. For best effect, do not use a ruler but draw it freehand as best you can.
1 In the middle of the page, draw a horizontal line AB approx 10 cm in length
2 At point A draw a perpendicular AC, about 6 to 7 cm in length.
3 At point B draw a line BD so that angle ABD is obviously larger than 90 deg. (Not too much, about 100 to 115 deg)
4 Mark off BD exactly the same length as AC.
5 Join C and D. (This line will slope slightly downwards relative to AB.)
6 Now draw a perpendicular bisector on both line AB and CD.
7 Since CD and AB are NOT parallel, these lines will cross somewhere. (The bisector of line CD will lie slightly to the right of the bisector of AB.)
8 Let the two bisectors cross at a point O in the middle of the lower half of the page. Join OA, OB, OC and OD.
9 Examine triangles OAC and OBD.
10 In these triangles AO = OB (O is equidistant from A and B)
CO = CD (O is equidistant from C and D)
AC = BD (By construction)
11 Hence, these triangles are congruent (three sides equal)
12 Hence, angle CAO equals angle OBD
13 But angle OAB equals angle OBA (isosceles triangle)
14 Hence, angle CAB (which is 90 deg) equals angle ABD (which is > 90)
Obviously, this cannot be true. Try to reason it out logically and mathematically, before being tempted to do the construction accurately, which will make the fallacy clear!
🙄 😲 🙄
This requires a pencil and ideally an A4 sheet of paper to do a construction. For best effect, do not use a ruler but draw it freehand as best you can.
1 In the middle of the page, draw a horizontal line AB approx 10 cm in length
2 At point A draw a perpendicular AC, about 6 to 7 cm in length.
3 At point B draw a line BD so that angle ABD is obviously larger than 90 deg. (Not too much, about 100 to 115 deg)
4 Mark off BD exactly the same length as AC.
5 Join C and D. (This line will slope slightly downwards relative to AB.)
6 Now draw a perpendicular bisector on both line AB and CD.
7 Since CD and AB are NOT parallel, these lines will cross somewhere. (The bisector of line CD will lie slightly to the right of the bisector of AB.)
8 Let the two bisectors cross at a point O in the middle of the lower half of the page. Join OA, OB, OC and OD.
9 Examine triangles OAC and OBD.
10 In these triangles AO = OB (O is equidistant from A and B)
CO = CD (O is equidistant from C and D)
AC = BD (By construction)
11 Hence, these triangles are congruent (three sides equal)
12 Hence, angle CAO equals angle OBD
13 But angle OAB equals angle OBA (isosceles triangle)
14 Hence, angle CAB (which is 90 deg) equals angle ABD (which is > 90)
Obviously, this cannot be true. Try to reason it out logically and mathematically, before being tempted to do the construction accurately, which will make the fallacy clear!
🙄 😲 🙄