884. COR. If two triedral angles have their diedral an gles equal each to each:

I. They are equal, when the equal diedral angles are arranged in the same order in both triedral angles.

II. They are symmetrical, when the equal diedral angles are arranged in opposite order in the two triedral angles.

885. The shortest line that can be drawn on the surface of a sphere between two points is the arc of a great circle, not greater than a semi-circumference, joining those points.

GIVEN an arc of a great circle AB, not greater than a semi-circumference, joining the points A and B on a spherical surface.

TO PROVE that AB is the shortest line that can be drawn on the surface between A and B.

Let C be any point of AB.

With A and B as poles describe circumferences whose polar distances are AC and BC.

These circumferences have only the point C in common. For, let D be any other point of the circumference whose pole is B.

Draw the great-circle-arcs AD and BD and let AD meet the circumference whose pole is A in E.

Therefore D lies outside the small circle whose pole is A, and the two small circles have only the point C in com

mon.

Now we will prove that the shortest line on the surface between A and B must pass through C.

Let AFGB be any line on the surface between A and B that does not pass through C.

It must cut the small circles in separate points F and G. Now, whatever may be the nature of the line AF, an equal line can be drawn on the surface between A and C.

[This can be shown by supposing the spherical surface to revolve on the axis of the small circle FCE, so that F will move along the small circle to C, while A remains fixed.]

Similarly a line equal to BG can be drawn from B to C. There will then lie between A and B and passing through Ca line less than AFGB by the portion FG.

We have now proved that through C can be drawn a line joining A and B less than any line joining A and B that does not pass through C.

Hence the shortest line must pass through C.

But C is any point in the arc AB.

Therefore the shortest line between A and B must pass through every point of the arc AB and hence must coincide with that arc.

CASE II. When AB is a semi-circumference.

Q. E. D.

We can show as above that any portion of the shortest line joining A and B must be an arc of a great circle, and that therefore the whole must be an arc of a great circle.

Q. E. D.

MEASUREMENT OF SPHERICAL FIGURES

886. Defs.-A lune is a portion of a spherical surface bounded by two semi-circumferences of great circles; as ACBDA.

The angle of a lune is the angle formed by its bounding

arcs.

Let P be the pole of the small circle passing through A, B, and C, and draw the great-circle-arcs PA, PB, and PC. Then

=

PA PB PC.

=

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Now place the two triangles vertically opposite to each other and draw the diameter POP'.

Also draw the great-circle-arcs P'A', P'B', and P'C'.

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The vertical triangles PBC and P'B'C' are symmetrical

and isosceles and therefore equal.

Similarly PCA = P'C'A' and PAB=P'A'B'.

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That is, the three parts of ABC are respectively equal to three parts of A'B'C'.

Therefore

area ABC-area A'B'C'.

Q. E. D.

888. COR. I. If two semi-circumferences of great circles BCB' and ACA' intersect on the surface of a hemisphere, the sum of the areas of the two opposite spherical triangles ABC and CA'B' is equal to the area of a lune whose angle is equal to BCA.

Hint.-Area ABC + area CA'B' area A'B'C' + area CA'B'.

889. COR. II. Two symmetrical spherical polygons are equivalent.

PROPOSITION XXXIV. THEOREM

890. Two lunes on the same sphere are equal, if their angles are equal.

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