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1. Four lines CA, CB, CD, CE meet at C; the angles ACB, BCD are equal and the angles DCE, ECA are equal; prove that BCE is a straight line.
2. Four lines CA, CB, CD, CE are such that the angles ACB, BCD are together equal to the angles DCE, ECA together; prove that ACD is a straight line.
3. ABC is an equilateral triangle and G is a point within it such that GA=GB=GC; D is the middle point of the side BC; prove that DGA is a straight line.
4. E is the middle point of the diagonal BD of the quadrilateral ABCD whose opposite sides are equal; prove that AE, EC is a straight line.
5. E is the middle point of the diagonal BD of the quadrilateral ABCD in which AB= AD and CB=CD; prove that AE, EC is a straight line.
6. If four straight lines meet at a point so that the opposite angles are equal each to each, then the four straight lines are two and two in the same straight line.
7. E is a point within the quadrilateral ABCD whose opposite sides are equal and EA = EC and EB=ED; prove that the straight lines AC, BD each pass through E.
8. ACB is a straight line; the lines CD, CE, CF, CG are drawn making angles DCA, ECA, DCB, ECB equal to each other; prove that the figure consists of three straight lines intersecting at C.
9. An even number of straight lines are drawn from the same point; prove that if the opposite angles are equal each to each, then opposite pairs of lines are in a straight line.
10. Two triangles ABC, DEF are such that AB=DE, the angles ABC, DEF are each a right angle and AC=DF; prove by placing the triangles so that AB and DE coincide 'back to back' that the triangles are equal in all respects.
11. Two triangles ABC, DEF are such that _BD=DE, AC=DF and the angles ABC, DEF together are equal to two right angles; prove, by the method of Question 10, that the angles BAC-EDF.
92. If two straight lines intersect, the vertically opposite angles at their point of intersection are equal.
Let the straight lines ACB, DCE intersect at C ;
it is required to prove
that the angles ACD, BCE are equal,
and that the angles ACE, BCD are equal.
Since ACB is a straight angle
and DCE is also a straight angle,
therefore the angles ACE, ECB together are equal to the angles DCA, ACE together; [Prop. 13.] taking away the common angle ACE, it follows that the angle ECB is equal to the angle DCA.
Similarly it may be shewn
that the angle ACE is equal to the angle DCB. Wherefore, if two straight lines intersect, etc.
1. The angle ACB is bisected by CD; AC, AD, AE are produced to E, F, G respectively; prove that CF bisects the angle ECG.
2. If in the figure of Prop. 15, AC=CB and DC=CE, then AD is equal to BE.
3. A and B are two given points on the same side of the given straight line DEGC; AE is drawn perpendicular to DC and AE is produced to F'so that FE=EA; FB is joined cutting DC in G; prove that AG and BG make equal angles with DC.
4. AG and BG make equal angles with the line DGC; the perpendicular from any point H in BG to the line DGC meets DG in E and AG in F; prove that HE=EF.
5. If the diagonals of a quadrilateral bisect each other the opposite sides of the quadrilateral are equal.
Since all straight angles are of the same size, all other angles are measured by means of a straight angle.
It is usual to divide a straight angle into 180 equal parts. One of these parts is called a degree.
[In like manner as the British trade standard of length, the yard, is divided into 36 equal parts, and each part is called an inch.
And the scientific standard of length, the metre, is divided into 100 equal parts, and each part is called a centimetre.]
Thus we shall prove that an angle of an equilateral triangle is one third of a straight angle
In the figure the angle BCE is about 10 degrees, the angle BCF is about 20 degrees.
PROPERTIES OF TRIANGLES.
94. When a side of a triangle is produced, the exterior angle is greater than either of the interior opposite angles.
Let a side BC of the triangle ABC be produced to D;
it is required to prove that the angle ACD
is greater than either the angle BAC or the angle ABC.
Let E be the middle point of AC.
In the straight line BE, produced beyond E,
Consider the triangles AEB, CEF;
the vertically opposite angles AEB, CEF are equal, [Prop. 15.] hence, because the two sides AE, EB and their included angle AEB in the triangle AEB, are equal to the two sides CE, EF and their included angle CEF, in the triangle CEF; therefore the two triangles are equal in all respects; [Prop. 4.] so that the angle BAE is equal to the corresponding angle ECF. But the angle ECD is greater than its part, the angle ECF, therefore the angle ECD is greater than the angle BAE.
Similarly, if G is a point in the side AC produced,
is greater than the corresponding interior angle ABC. But ACD and BCG, being vertically opposite angles, are equal;
therefore also ACD is greater than the angle ABC. Wherefore, when a side of a triangle is produced, etc.
Example. If two equal angles have one arm coincident and in the same direction then their other arms cannot intersect unless they are coincident.
Let ABC, DEF be two equal angles and let the arms BC and EF be coincident and in the same direction; it is required to prove that BA and ED cannot have one point only common.
For suppose a point G only were common to both lines, then GBE would be a triangle and one of the two angles ABC, DEF must be an exterior angle and the other an interior opposite angle of the triangle GBE in which case the angles ABC, DEF are unequal by Prop. 16. So that BA, DE cannot be sides of a triangle. Q.E.D.