12 I. ON ANGLES
O
12
3 4
Figure 11
Theorem. In a given plane, through a given point on a line, one can draw
one, and only one, perpendicular to this line.
Suppose we want to draw a perpendicular to line AA through point O (Fig.
11). It suffices to draw a circle with center O which cuts the line at A and A , then
find point B which divides semicircle AA into two equal parts. Then OB will be
the required perpendicular; and conversely, a perpendicular to AA drawn through
point O must divide arc AA in half.
Corollary. We see that a right angle intercepts an arc equal to one fourth of
a circle centered at the vertex of the angle.
Figure 12
All right angles are equal, since two such angles will intercept equal arcs on
equal circles centered at their vertices.
15. If several rays are drawn issuing from the same point, the sum of the
successive angles thus formed (AOB, BOC, COD, DOA, Fig. 12) is equal to four
right angles.
Indeed, the sum of the arcs intercepted by these angles on a circle centered at
their vertex is the entire circle.
If several rays are drawn issuing from a point on a line, in such a way that all
the rays are on one side of the line (Fig. 13), the sum of the successive angles thus
formed is equal to two right angles.
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