Elements of Plane Geometry According to Euclid

mn, and EBC næ; also angle mn=m+n and nx= n+a; therefore

m n + x

=m+n+ x;

but m+nx = m + n + x;

therefore m n +x and m+nx, being equal to the same three angles, are equal to one another; that is, mn+x=m+nx = two right angles.

PROPOSITION XIV. THEOREM.

If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD, equal together to

two right angles. BD is in the same straight line with CB.

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it; then, because the straight line AB makes angles with the straight line CBE, upon one side of it, therefore the angles ABC, ABE, are together equal (I. 13) to two right angles; but the angles ABC, ABD, are likewise together equal to two right angles; therefore the angles CBA, ABE, are equal to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is equal (Ax. 3) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB.

PROPOSITION XV. THEOREM.

If two straight lines cut one another, the vertical or opposite angles shall be equal.

Let the two straight lines AB, CD, cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.

For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (1.13) to two right angles; and the angles c AED, DEB, which the straight line a DE makes with the straight line AB,

are also together equal to two right angles; therefore the two angles CEA, AED, are equal to the two AED, DEB. Take away the common angle AED, and the remaining angle CEA is equal (Ax. 3) to the remaining angle DEB. In the same manner it can be demonstrated that the angles CEB, AED, are equal.

COR. 1.-From this it is manifest, that, if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles.

COR. 2. And hence, all the angles made by any number of lines meeting in one point, are together equal to four right angles.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAC.

Bisect (I. 10) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.

Because AE is equal to EC, and BE to EF; AE, EB, are equal to CE, EF, each to each; and the angle AEB is equal to the angle CEF (I. 15), because they are opposite vertical angles; therefore (1.4) the

base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF ; therefore the angle ECD, that is, ACD, is greater than BAE.

mn, and EBC na; also angle mn=m+n and nx= n+x; therefore

mn + x = m + n + x;

but m+ n x = m + n + x;

therefore m n +x and m+nx, being equal to the same three angles, are equal to one another; that is, mn + x = m + nx = two right angles.

PROPOSITION XIV. THEOREM.

At the point B in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD, equal together to two right angles. BD is in the same straight line with CB.

For, if BD be not in the same straight line

with CB, let BE be in the same straight line C B with it; then, because the straight line AB makes angles with the straight line CBE, upon one side of it, therefore the angles ABC, ABE, are together equal (I. 13) to two right angles; but the angles ABC, ABD, are likewise together equal to two right angles; therefore the angles CBA, ABE, are equal to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is equal (Ax. 3) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB.

PROPOSITION XV. THEOREM.

If two straight lines cut one another, the vertical or opposite angles shall be equal.

Let the two straight lines AB, CD, cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.

For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (1.13) to two right angles; and the angles c. AED, DEB, which the straight line a DE makes with the straight line AB,

COR. 1.-From this it is manifest, that, if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles.

COR. 2. And hence, all the angles made by any number of lines meeting in one point, are together equal to four right angles.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAC.

Bisect (1.10) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.

Because AE is equal to EC, and BE to EF; AE, EB, are equal to CE, EF, each to each; and the angle AEB is equal to the angle CEF (I. 15), because they are opposite vertical angles; therefore (1.4) the base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is, ACD, is greater than BAE.

In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is (I. 15), the angle ACD, is greater than the angle ABC.

PROPOSITION XVII. THEOREM.

Any two angles of a triangle are together less than two right angles.

Let ABC be any triangle; any two of its angles together are less than two right angles.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC (I. 16); to each of these add the angle ACB; therefore the angles ACD, ACB, are greater than the angles ABC, ACB; but ACD, ACB, are together equal to two right angles (I. 13); therefore the angles ABC, BČA

are less than two right angles. In like manner it may demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles.

The greater side of every triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA.

Because AC is greater than AB, make (I. 3) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DCB (I. 16); but ADB is equal

to ABD (I. 5), because the side AB is equal to the side AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore the angle ABC is still greater than ACB.

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