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Cor. 1.—The sum of the interior angles of any rectilineal figure is equal to twice as many right angles as the figure has sides, minus four right angles. Take any rectilineal figure, as
B ABCDEF, and take G, any point within it. Join G with A, B, etc. (1.) There are as many triangles as
A the figure has sides.
(2.) The angles in each triangle amount to two right angles (I. 32). (3.) .. the angles in all the triangles
F together amount to twice as many right angles as the figure bas sides (2n L.S., n=number of sides).
(4.) But the angles of all the triangles include the angles of the figure at A, B, C, etc., and the angles at G which amount to four right angles.
(5.) ... the angles of the figure amount to twice as many right angles as the figure has sides, minus four right angles (2n L^.s. — 4 L.S.)
Cor. 2.—The sum of the exterior angles of any convex rectilineal figure, formed by producing the sides all in one way, is four right angles.
Take ABCDEF, any convex rectilineal figure having the sides produced all in one way.
To prove sum of exterior angles four right angles.
•G (1.) Each exterior angle with its adjacent interior angle, as 21+ < 2, equals two right angles.
(2.) :. all the exterior with all the interior amount to twice as many right angles as there are sides in the figure (2n L.s.)
(3.) But all the interior angles together with the angles at G amount to twice as many right angles as the figure has sides (Cor. 1),(2n L'.s.)
(4.) .. all the exterior angles with all the interior =all the interior angles with the angles at G.
(5.) .. all the exterior angles together= angles at G=four right angles (4 ll.s.)
Note.—When the rectilineal figure is not convex, but has a reflex angle or reflex angles, the result is modified as follows; the proof of which we leave to the student:
21+22+ L3+2 4+ 25-26=
80. The angles of a quadrilateral amount to four right angles.
81. Find the magnitude of the angle in a regular pentagon, hexagon, heptagon, octagon, etc. (Cor. 1).
82. Find the magnitude of the exterior angle of any regular polygon, as a pentagon, nonagon, decagon, or polygon of n sides.
83. The exterior angle of a regular polygon is ts of a rt. angle, how many sides has it ?
84. The interior angle of a regular polygon is 1f rt. angle, how many sides has it ?
85. Show that three regular hexagons can be placed so as to have a common angular point and to fill up the space about it.
86. What other regular figures can be formed round common angular points in the same way?
87. Make a pattern composed of octagons and squares.
88. What other regular figures may be combined to fill up the space round the cominon angular points ? Make patterns of the combinations.
89. How many diagonals can be drawn in a pentagon ; in a hexagon; in a polygon of n sides?
90. If all the sides of a polygon be produced in both ways, the angles formed by the meeting of every two alternate sides will together amount to twice as many right angles as the figure has sides, minus eight right angles (2n Ls.
-8 L1.8.) 91. The middle point of the hypotenuse of a right-angled triangle is equally distant from each of the three angles.
92. And conversely, If the middle point of one side of a triangle be equally distant from each of the three angles, the triangle shall be right-angled.
93. Given the perimeter of a triangle and the angles at the base, to construct the triangle.
PROP. 33. THEOR.
The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are also themselves equal and parallel. Given AB = and || CD, and
let them be joined towards
the same parts. To prove AC = and || BD. Join BC. .: BC meets the parallels AB, CD, :: L ABC
_ BCD (1. 29), and AB CD (given), and BC is common, .:. Δ ABC
ABCD in every respect (I. 4)
:. AC = BD. Also L ACB = CBD, and .:: AC is also || BD (1. 27).
Note.- Propositions have already occurred where the hypothesis and conclusion of one become the conclusion and hypothesis of another, as Props. 5 and 6; such propositions are converse the one of the other.
94. Mention some examples of converse propositions. Is the converse of a true proposition always a true proposition, e.g., the converse of " All squares are parallelograms ?” Enunciate a converse of Ex. 68, of Ex. 70, of Ex. 77, and of Prop. 33.
95. The straight lines joining the extremities of two equal and parallel straight lines towards the opposite parts bisect each other.
96. A square, rectangle, rhombus, and rhomboid are all parallelograms (Def. 29).
97. The straight lines joining the extremities of two lines which bisect each other form a parallelogram.
98. If lines are respectively parallel to, or respectively perpendicular to, the sides of a triangle, they shall form a triangle equiangular to the given triangle.
PROP. 34. THEOR.
i.e., given AB || DC, and
DC, AD BC, 2D B, LDAB = _ DCB, and A ADC A ABC. :: AB is || DC, and AC meets them, :: BAC = L ACD (I. 29); :: AD is || BC, and AC meets them, :: LDAC
= L ACB (I. 29); and AC is common to triangles ADC, ABC; :: AADC = A ABC (I. 26), :. the diagonal bisects the
parallelogram; and :. also AB = DC, AD BC, and 2D = 2B; and · DAB _ DCB, : _ BAC and 2 DAC are
respectively equal to L ACD and L ACB. Q.E.D.
99. The diagonals of a parallelogram bisect each other.
100. The line joining the middle points of two sides of a triangle is parallel to the base, and equal to half the base of the triangle.
101. The lines joining the middle points of the adjacent sides of a quadrilateral form a parallelogram.
102. If a straight line be drawn parallel to the base of a triangle, bisecting one side, it shall also bisect the other side.
103. The diagonals of a square or a rhombus bisect each other at right angles.
104. If the opposite sides or opposite angles of a quadrilateral are equal, the figure is a parallelograin.
On Equality of Area of certain Rectilineal
PROP, 35. THEOR. Parallelograms on the same base and between the same parallels are equal to one another.
In Fig. 1, Let ABCD, DBCE be the parallelograms on
base BC, and between the parallels BC, AE. To prove parallelograms AC and BE equal. :: | AC is double A DBC (1. 34), and BE is also double A DBC (1. 34),
:: 0 AC = | BE (Ax. 6). In Fig. 2 and Fig. 3, Let ABCD and EBCF be the
parallelograms. In triangles EAB and FDC, :: AB = DC (1. 34), and LA = FDC (1. 29), and 2 AEB = LF (I. 29),
:: AEAB= AFDC (1. 26). Now let AEAB be taken from the figure ABCF, and
| BF remains. Replace A EAB, and take a FDC from the figure ABCF, and [ AC remains.
:: BF = | AC (Ax. 3). Q.E.D. Note.—This proposition is the first of a series which prove that certain figures may have equal areas without being congruent.
105. In I. 85, prove A EAB = A FDC by I. 4; also by I. 8.