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Fwd: Re: Saccheri quadrilaterals
Posted:
Jun 6, 2007 7:57 PM


I'm trying to picture your drawing... Does the quadrilateral inside the other quadrilateral share a common bottom base (with the two right angles) with the arms of the smaller lying on the arms of the larger? In that case you have two quadrilaterals. The lower one is a saccheri quadrilateral but the top one is not. It has two obtuse angles and two acute angles.
I can see how case 2 is a contradiction, and cannot occur. Case 3 can occur. However, they cannot fit one inside the other to make two quadrilaterals. If they share a common baseline and perpendicular bisectors to the bases, then the arms of one will cross the upper base (summit) of the other.
>To: geometrycollege@support1.mathforum.org > >I was trying to do this too. And this is what I think should work to >prove the statement > >Assume the saccheri quadrilaterals are not congruent. That means for >these two quadrilaterals, > >There are three cases where the quadrilaterals will not be equal: > >1. arms are not equal and bases are not equal >2. arms are not equal and bases are equal >3. arms are equal and bases are not equal > >If can be proven easily that when arms are equals base must be >equal. Thus 3 is a contradiction. > >When arms are not equal, assume one of the quadrilateral arms are >larger than the other. Then in the larger the quadrilateral you can >construct another quadrilateral (inside) where arms are equal to the >smaller quadrilateral's arms. You can show this inside quadrilateral >is congruent to the smaller quadrilateral. But that is not possible >as it would create a leftover quadrilateral with anglesum larger than 360. > >Let me know if you got a different way to proving this. > >Sa'ad >mashrur.mia@gmail.com
Ben Saucer email: bsaucer2@comcast.net web page: www.saucersdomain.com ICQ: 20610314



