Logic Q

[Amina-DZ]

How to prove a right angle congruent to an obtuse angle?



Warning: False proof ahead. An obtuse angle is certainly not congruent to a right angle. Can you find the well-hidden fallacy? Maybe you should not trust what you see!

Consider the above picture: ABCD is a rectangle and E is a point near D slightly outside the rectangle so that AE is equal to AD. H is the middle of CD and K is the middle of CE. The perpendicular to CD going through H and the perpendicular to CE going though K intersect at a certain point J.

Now consider the sides of the triangles BCJ and AEJ: First, BC=AE (since both of these are equal to AD). Second, JB=JA (J is on the perpendicular bisector of CD, which is also that of AB). Third, CJ=EJ (J is on the perpendicular bisector of CE, by construction). The inescapable conclusion (I swear it's true!) is that BCJ and AEJ are congruent triangles (as their 3 sides are congruent), therefore the angles CBJ and EAJ are equal...

The angles JBA and JAB are equal (since JAB is an isosceles triangle), the picture clearly tells you that if you subtract JBA from CBJ and and JAB from JAE you obtain ABC and BAE. Yet one of these (ABC) is a right angle, whereas the other (BAE) is obtuse by construction. How can this be?
HINT: Every statement is true,
but there are misleading words in the last paragraph.

The American mathematician Paul R. Halmos (born in Budapest on March 3, 1916) rightly said: "You are allowed to lie a little, but you should never mislead". Here, the challenge is to find a misleading statement which is not a lie...



P.S. You should be able to see the pic! Oh and good luck!

[Shotokan_Karate]

Oh no! Did I speak too soon! I can see I'm gonna have fun tonight No promises though!

[Amina-DZ]

Originally posted by Shotokan_Karate

Oh no! Did I speak too soon! I can see I'm gonna have fun tonight No promises though!

Be careful what you wish for!

[Shotokan_Karate]

So good job I wished for an easy puzzle Well I did say be gentle

mmm...I think it's not as difficult as it looks Let's see...

The key to solving this kind of puzzles is to redraw the shapes accurately. Remember the magic triangle

Here I drew the green circle to help me out.



We can clearly see that there is a difference in shape to the one provided.

Now we go through the statements again, so...

Consider the above picture: ABCD is a rectangle and E is a point near D slightly outside the rectangle so that AE is equal to AD. H is the middle of CD and K is the middle of CE. The perpendicular to CD going through H and the perpendicular to CE going though K intersect at a certain point J.

...Fine

Now consider the sides of the triangles BCJ and AEJ: First, BC=AE (since both of these are equal to AD). Second, JB=JA (J is on the perpendicular bisector of CD, which is also that of AB). Third, CJ=EJ (J is on the perpendicular bisector of CE, by construction). The inescapable conclusion (I swear it's true!) is that BCJ and AEJ are congruent triangles (as their 3 sides are congruent), therefore the angles CBJ and EAJ are equal...

...Fine

The angles JBA and JAB are equal (since JAB is an isosceles triangle), the picture clearly tells you that if you subtract JBA from CBJ and and JAB from JAE you obtain ABC and BAE. Yet one of these (ABC) is a right angle, whereas the other (BAE) is obtuse by construction. How can this be?

...uhuh...I don't think so! True that "the picture clearly tells us that..." but the picture provided is not right! The line JE should be outside the angle AJD! And thus one cannot subtract JAB from JAE as claimed which in turn annuls the statement that a right angle is congruent to an obtuse angle.

no?

[Amina-DZ]

you figured it out!!

I know! It's because I was...ahem gentle!

Wait and you'll see

[Shotokan_Karate]

I can't wait to see

[Amina-DZ]

Let's see, I'm still being gentle

You leave home, you walk one kilometer south, then one kilometer due east and finally one kilometer north. If you're home again at the end of your walk, where's home?