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For the angles CDA and DCE, thus formed equal, are the alternate angles which CD makes with the straight lines CE and AB, and, therefore, by the corollary to the last proposition, these lines are parallel.
PROP. XXIV. THEOR.
Parallel lines are equidistant, and equidistant straight lines are parallel. -
The perpendiculars EG, FH, let fall from any points E, F in the straight line AB upon its parallel CD, are equal ; and if these perpendiculars be equal, the straight lines AB and CD are parallel.
For join EH: and because each of the interior angles EGH and FHG is a right angle, they are together equal to two right angles, and consequently the perpendiculars EG and FH are (I. 22. cor.) parallel to each other; wherefore (I. 22.) the alternate angles HEG and EHF are equal. c-AF II D But, EF being parallel to GH, the alternate angles EHG and HEF are likewise equal; and thus the two triangles HGE and HFE, having the angles HEG and EHG respectively equal to EHF and HEF, and the side EH common to both, are (I.20.) equal, and hence the side EG is equal to FH.
Again, if the perpendiculars EG and FH be equal, the two triangles EGH and EFH, having the side EG equal to FH, EH common, and the contained angle HEG equal to EHF, are (I. 3.) equal, and therefore the angle EHG equal to HEF, and (I. 22.) the straight line AB parallel to CD. - ** :
PROP. XXV. THEOR. **:1
The opposite sides of a rhomboid are parallel.
If the opposite sides AB, DC, and AD, BC of the quadrilateral figure ABCD be equal, they are also parallel.
For draw the diagonal AC. And because AB is equal to DC, BC to AD, and AC is common; the two triangles ABC and ADC are (I. 2.) equal. Consequently the angle ACD is equal P C to CAB, and therefore the side AB (I, 22. cor.) parallel to CD; and, for the same reason, the angle CAD being equal to ACB, the side AD is parallel to BC.
Cor. Hence the angles of a square or rectangle are all of them right angles; for the opposite sides being equal, are parallel; and if the angle at A be right, the other interior one at B is also a right angle (I. 22.), and consequently the angles at C and D, opposite to these, are right.—On this proposition depends the construction of the instrument called a Parallel Ruler.
PROP. XXVI. THEOR.
The opposite sides and angles of a parallelogram are equal.
Let the quadrilateral figure ABCD have the sides AB and BC parallel to CD and AD; these are respectively equal, and so are likewise the opposite angles at A and C, and at B and D.
For join AC. Because AB is parallel to CD, the alternate angles BAC and ACD are (I. 22.) equal; and since AD is parallel to BC, the alternate angles ACB and CAD are also equal. Wherefore the triangles ABC and ADC, having the angles CAB and ACB equal to ACD and CAD, and the P C interjacent side AC common to both, are (I.20.) equal. Consequently, the side AB is equal to CD, and the side BC to AD; and these opposite sides being thus equal, the opposite angles (I. 25.) must be likewise equal.
Cor. Hence the diagonal divides a rhomboid or parallelogram into two equal triangles.
PROP. XXVII. THEOR.
If the parallel sides of a trapezoid be equal, the other sides are likewise equal and parallel.
Let the sides AB and DC be equal and parallel; the
sides AD and BC are themselves equal and parallel. For draw the diagonal AC. Because AB is parallel to
CD, the alternate angles CAB and ACD are (I. 22.) equal; and the triangles ABC and ADC, having the side AB equal to CD, AC common to both, B and the contained angle CAB equal to ACD, are, therefore, equal (I. 3.). Whence the side BC is equal to AD, * C and the angle ACB equal to CAD; but these angles being alternate, BC must also be parallel to AD (I. 22. cor.)
PROP. XXVIII. THEOR.
Lines parallel to the same straight line, are pa
rallel to each other.
If the straight line AB be parallel to CD, and CD pa
rallel to EF; then is AB parallel to EF. For let a straight line GH cut these :
fore the angle GIA is equal to GLE, and consequently
AB is parallel to EF (I. 22. cor.)
PROP. XXIX. THEOR.
Straight lines drawn parallel to the sides of an
angle, contain an equal angle.
If the straight lines AB, AC be parallel to DE, DF; the angle BAC is equal to EDF. For draw the straight line GAD through the vertices. And since AC is parallel to DF, the exterior angle GAC is (I, 22.) equal to GDF; and,
for the same reason, GAB is equal to GDE; there con
sequently remains the angle BAC equal to EDF.
PROP. XXX. THEOR.
An exterior angle of a triangle is equal to both its opposite interior angles, and all the interior angles of a triangle are together equal to two right angles.
The exterior angle BCD, formed by the production of the side AC of the triangle ABC, is equal to the two opposite interior angles CAB and CBA, and all the interior angles CAB, CBA and BCA of the triangles are together equal to two right angles. For, through the point C, draw (I. 23.) the straight line CE parallel to AB. And, AB being parallel to CE, the interior angle BAC is (I.22.) equal to the exterior one ECD ; and, for the same reason, the alternate angle ABC is equal to BCE. Wherefore the two angles, CAB and ABC are equal to TB E. DCE and ECB, or to the whole exterior angle BCD. Again, add the adjacent angle BCA A C ID to the exterior angle BCD, and to the two interior angles CAB and ABC; and all the interior angles of the triangle ABC are together equal to the angles BCD and BCA on the same side of the straight line AD, that is, to two right angles. Cor. 1. Hence the two acute angles of a right angled triangle are together equal to one right angle; and hence each angle of an equilateral triangle is two-third parts of a right angle. Cor. 2. Hence if a triangle have its exterior angle, and one of its opposite interior angles, double of those in an