Saccheri quadrilaterals

[Swartz]

Hi there.

I am trying to prove that two Saccheri quadrilaterals are congruent to one
another if their summits and summit angles are equal.

For example,
Saccheri quadrilaterals ABCD, WXYZ with bases AB and WX respectively.
Given summit CD = YZ. Angle at C = D = Y = Z. How do you prove that ABCD is
congruent to WXYZ?

Any hints are welcome?

Thanks in advance.

[Mashrur Mia]

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 left-over quadrilateral with angle-sum larger than 360.

Let me know if you got a different way to proving this.

Sa'ad
mashr...@gmail.com

[Mashrur Mia]

Opps sorry I said

> But that is not possible as it would create a left-over
> quadrilateral with angle-sum larger than 360.

It should be

But that is not possible as it would create a left-over

quadrilateral with angle-sum equals to 360.

and thus it contradicts 1 and 2

[K. E. Pledger]

> ....

> Saccheri quadrilaterals ABCD, WXYZ with bases AB and
> WX respectively.
> Given summit CD = YZ. Angle at C = D = Y = Z. How do
> you prove that ABCD is

> congruent to WXYZ? ....

Mashrur Mia has suggested matching the two quadrilaterals at their bases, but it seems more efficient (and avoids splitting cases) to match them at their summits.

Suppose AD = BC > WZ = XY.
Let U be the point on BC such that CU = YX,
and let V be the point on AD such that DV = ZW.

The quadrilaterals VUCD and WXYZ can be proved congruent (by introducing diagonals CV, YW, and using two pairs of congruent triangles). But this gives right angles at U and V, so the quadrilateral ABUV has angle sum 2(pi), which is too much.

A similar contradiction arises from AD < WZ; so AD = WZ, etc.

Ken Pledger.

[BSK]

> Hi there.
>
> I am trying to prove that two Saccheri quadrilaterals
> are congruent to one
> another if their summits and summit angles are equal.
>
> For example,
> Saccheri quadrilaterals ABCD, WXYZ with bases AB and
> WX respectively.
> Given summit CD = YZ. Angle at C = D = Y = Z. How do
> you prove that ABCD is
> congruent to WXYZ?

They are not congruent.

[chasjac too]

BSK wrote:

I assume that the unstated axiom here is that the geometry is non-Euclidean,
in which case Ken Pledger's argument is correct -- you cannot have a
rectangle on the hyperbolic plane or the sphere. So, under the stated
conditions, this is a congruence theorem for Saccheri quads in the presence
of some negation to EPP.

But this argument would not hold in the Euclidean plane. So, there's an
equivalence theorem hiding here.

The following are equivalent:

(1) Two Saccheri quads are congruent whenever their summits and summit
angles are equivalent;

(2) There are no rectangles.

proof outline: (2) => (1) is what Ken did. Do (1) => (2) by
contrapositive.

--charlie