Euclidean and Non Euclidean Geometry

Problem 9.1 Side-Side-Side (SSS)

Suppose a triangle in figure:

If we line up sets of comparing sides of the two triangles, we have two unique directions for

different sets of sides.

Each of these directions is conceivable, and to demonstrate or negate SSS. Once More, balance

of triangle can be exceptionally valuable here.

On a circle, SSS does not work for all triangles. The counterexample in underneath figure

shows that regardless of how little the sides of the triangle are. SSS does not hold on the

grounds that all the three sides consistently decide two unique triangles on a circle.

Let ABC be a triangle. Round’s law of cosine permits us to settle for the cosines of ∠A, ∠B, ∠C

utilizing only the lengths of the sides. Since these points lie carefully among 0 and π, the cosines

decide the points. Accordingly, we can recuperate ∠A, ∠B, and ∠C simply regarding the lengths

of the sides.

In specific, any two triangles with equivalent sides would likewise have equivalent points, and

would be accordingly compatible as demonstrated beneath:

Problem 9.2 Angle-Side-Side (ASS)

Suppose you have two triangles with the above congruencies. We will call them ASS triangles.

We should check whether, truth is told, the triangles are steady.

We can mastermind the point and the fundamental side, and we know the length of the

ensuing side, yet we don't have even the remotest clue where the second and third sides will

meet. See Figure:

Here, the circle that has as its reach the second side of the triangle meets the bar that goes

along the direct path from “a” to “b” twice. So, ASS doesn't work for all triangles on either the

plane or a circle or an overstated plane.

At that point, it is inferred that RLH is valid for little triangles on a circle. However, there exists little triangle counter examples to RLH on circles!

The counterexample will assist you with seeing a few manners by which circles are characteristically altogether different from the plane.

The second leg of the triangle meets the geodesic that contains the third side a limitless number of times. So, on a circle there are little triangles which fulfill the states of RLH despite the fact that they are non-compatible.

However, in the event that you take a gander at your contention for RLH on the plane, you ought to have the option to show that on a circle, RLH is legitimate for a triangle with all the sides under 1/4th part of an extraordinary circle.

RLH is likewise valid for a bigger assortment of triangles on a circle.

Problem 10.1 Parallel Transport on the Plane

Fig: Parallel transport on a sphere along l, but not along l'

Fig: Parallel transport on a hyperbolic plane along l but not along l'.

Parallel Circles on a Sphere

The scope of the circles on the earth is here and is called "Equals of Latitude." They are equal as they are equidistant and are concentric circles on the plane. By and large, cross-over do not cut equidistant circles at compatible points. Nonetheless, there is one significant situation where cross-over do cut the circles at consistent points. Let l and l’ be scope circles which are a similar separation from the equator on inverse sides of it. At that point, each point on the equator is a focal point of half turn evenness for each pair of scope. Accordingly, every cross-over cuts these scope circles in congruent angles.

Figure: Special equidistant circles.

Parallel Postulates

It expresses that, in two-dimensional math: If a line portion converges two straight lines shaping two inside points on similar side that whole to under two right points, at that point; the two lines, whenever broadened uncertainly, meet on that side on which the points total to under two right points.

On a circle, all straight lines converge twice which implies that EFP is inconsequentially obvious. EFP is bogus on an exaggerated plane.

Figure: Euclid's Parallel Postulate.

Accordingly, EFP does not need to be accepted on a circle, it tends to be demonstrated! Notwithstanding, in most calculation course books, EFP is subbed by another propose which, it is guaranteed, is identical to EFP. This hypothesize is Playfair's Parallel Postulate (PPP), and it very well may be communicated in the accompanying manner: For each line l and each point P not on l, there is an exceptional line l' which goes through P and is corresponding to l.

Figure: Playfair’s Parallel Postulate.

It is to be noted that on a circle, any two incredible circles converge, there are no lines which

are corresponding to l in the "not meeting" sense. Consequently, Playfair's Postulate isn't

accurate on circles. Then again, on the off chance that we change "equal" to "equal vehicle"

each incredible circle through “P” is an equal vehicle of l along some cross-over.