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THEOREM 3. [Euclid I. 15.] If two straight lines cut one another, the vertically opposite angles are equal.

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Let the straight lines AB, CD cut one another at the point O.
It is required to prove that

(i) the L AOC = the 2 DOB;
(ii) the L COB= the L AOD.

Proof. Because AO meets the straight line CD,
... the adjacent LAOC, AOO together = two right angles;

that is, the L AOC is the supplement of the 2 AOD.

Again, because DO meets the straight line AB, .:. the adjacent , DOB, AOD together = two right angles ;

that is, the Z DOB is the supplement of the LAOD. Thus each of the 4* AOC, DOB is the supplement of the L AOD,

.. the L AOC = the DOB. Similarly, the - COB= the LAOD.

Q.E.D.

PROOF BY ROTATION.

Suppose the line COD to revolve about O until OC turns into the position o Then at the same moment OD must reach the position OB (for AOB and COD are straight).

Thus the same amount of turning is required to close the LAOC as to close the LDOB.

: the LAOC=the LDOB.

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EXERCISES ON ANGLES.

(Numerical.) 1. Through what angles does the minute-hand of a clock turn in (i) 5 minutes, (ii) 21 minutes, (iii) 433 minutes, (iv) 14 min. 10 sec.? And how long will it take to turn through (v) 66°, (vi) 222° ?

2. A clock is started at noon : through what angles will the hour. hand have turned by (i) 3.45, (ii) 10 minutes past 5? And what will be the time when it has turned through 1721° ?

3. The earth makes a complete revolution about its axis in 24 hours. Through what angle will it turn in 3 hrs. 20 min., and how long will it take to turn through 130°?

4. In the diagram of Theorem 3

(i) If the LAOC=35°, write down (without measurement) the value of each of the 28 COB, BOD, DOA.

(ii) If the 4* COB, AOD together make up 250°, find each of the LOCOA, BOD.

(iii) If the 28 AOC, COB, BOD together make up 274°, find each of the four angles at O.

(Theoretical.) 5. If from O a point in AB two straight lines OC, OD are drawn on opposite sides of 3 so as to make the angle COB equal to the angle AOD; shew that OC and OD are in the same straight line.

6. Two straight lines AB, CD cross at O. If OX is the bisector of the angle BOD, prove that XO produced bisects the angle AOC.

7. Two straight lines AB, CD cross at 0. If the angle BOD is bisected by Ox, and AOC by OY, prove that OX, OY are in the same straight line.

8. If OX bisects an angle AOB, shew that, by folding the diagram about the bisector, OA may be made to coincide with OB. How would OA fall with regard to OB, if

(i) the LAOX were greater than the LXOB;

(ii) the LAOX were less than the LXOB? 9. AB and CD are straight lines intersecting at right angles at O; shew by folding the figure about AB, that OC may be made to fall along OD.

10. A straight line AOB is drawn on paper, which is then folded about O, so as to make OA fall along OB ; shew that the crease left in the paper is perpendicular to AB.

ON TRIANGLES

1. Any portion of a plane surface bounded by one or more lines is called a plane figure.

The sum of the bounding lines is called the perimeter of the figure. The amount of surface enclosed by the perimeter is called the area.

2. Rectilineal figures are those which are bounded by straight lines.

3. A triangle is a plane figure bounded by three straight lines.

4. A quadrilateral is a plane figure bounded by four straight lines.

5. A polygon is a plane figure bounded by more than four straight lines.

6. A rectilineal figure is said to be

equilateral, when all its sides are equal;
equiangular, when all its angles are equal;
regular, when it is both equilateral and equiangular

7. Triangles are thus classified with regard to their sides A triangle is said to be

equilateral, when all its sides are equal ;
isosceles, when two of its sides are equal ;
scalene, when its sides are all unequal.

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In a triangle ABC, the letters A, B, C often denote the magnitude of the several angles (as measured in degrees); and the letters a, b, c the lengths of the opposite sides (as measured in inches, centimetres, or some other unit of length).

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Any one of the angular points of a triangle may be regarded as its vertex; and the opposite side is then called the base.

In an isosceles triangle the term vertex is usually applied to the point at which the equal sides intersect; and the vertical angle is the angle included by them.

8. Triangles are thus classified with regard to their angles : A triangle is said to be right-angled, when one of its angles is a right angle; obtuse-angled, when one of its angles is obtuse; acute-angled, when all three of its angles are acute. [It will be seen hereafter (Theorem 8. Cor. 1) that every triangle must have at least two acute angles.]

Right-angled Triangle. Obtuse-angled Triangle. Acute-angled Triangle.

In a right-angled triangle the side opposite to the right angle is called the hypotenuse.

9. In any triangle the straight line joining a vertex to the middle point of the opposite side is called a median.

THE COMPARISON OF TWO TRIANGLES.

(i) The three sides and three angles of a triangle are called its six parts. A triangle may also be considered with regard to its area.

(ii) Two triangles are said to be equal in all respects, when one may be so placed upon the other as to exactly coincide with it; in which case each part of the first triangle is equal to the corresponding part (namely that with which it coincides) of the other; and the triangles are equal in area

In two such triangles corresponding sides are opposite to equal angles, and corresponding angles are opposite to equal sides.

Triangles which may thus be made to coincide by superposition are said to be identically equal or congruent

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THEOREM 4. [Euclid I. 4.] If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, , then the triangles are equal in all respects.

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Let ABC, DEF be two triangles in which

AB=DE,

AC=DF, and the included angle BAC = the included angle EDF. It is required to prove that the ABC = the A DEF in all respects.

Proof. Apply the A ABC to the A DEF,

so that the point A falls on the point D,
and the side AB along the side DE.

Then because AB=DE,
.:. the point B must coincide with the point E.

And because AB falls along DE,

and the 2 BAC = the LEDF,
::: AC must fall along DF.

And because AC=DF,
.. the point C must coincide with the point F.
Then since B coincides with E, and C with F,
... the side BC must coincide with the side EF.
Hence the ABC coincides with the A DEF,
and is therefore ecual to it in all respects.

Q.E.D.

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