2. If two straight lines intersect and one of the angles formed is a right angle, the other three angles are also right angles.

Let A B and C D be two straight lines intersecting in O, and let AOC be a right angle; then the other angles at O are right angles.

A

B

Because one of the adjacent angles COA, COB, namely COA, is a right angle; ... by the definition of a right angle C O B is also a right angle.

Because BOD and AOC are vertically-opposite angles they are equal; ... B O D is a right angle;

Because A O D and C O B are vertically-opposite angles they are equal; .. A OD is a right angle.

3. The bisectors of two vertically-opposite angles are in the same straight line.

Let OE bisect the angle C O A, and let O F bisect

the vertically-opposite angle BOD;

then E F is a straight line.

I

Because COE = COA and

DOF=DOB,

2

2

.. <COE=<DOF;

But the continuation of E O makes with OD an angle equal to its vertically-opposite angle CO E. ..OF is the continuation of E O and E F is a straight line.

4. From a given point on a straight line two straight lines, and only two, can be drawn, making with it a given angle.

E

Let AB be the given straight line and C the point in it. First, let the given angle be less than a right angle. Draw CE perpendicular to AB; let a line, C D, revolve about the point C, and let it first coincide with C A. At first the angle A CD is zero, and therefore less than the given angle, and it gradually increases as CD revolves until CD coincides with CE. There must, therefore, be some position of C D between CA and C E such that ACD the given angle.

A

B

After CD has passed CE, 4 BCD is at first greater than and finally less than the given angle. There must, therefore, be some position of CD between C E and C B such that BCD is equal to the given angle.

If the angle be greater than a right angle we must consider D C B in the first half and D C A in the second half of the revolution.

5. The sum of all the angles formed by a number of straight lines that meet in a point is equal to four right angles.

Let A B, AC, AD, &c., be straight lines from the same point, A.

A

right angles.

Continue one of them, BA, along A E; then all the angles B above EB make up the two adjacent angles CA B, CA E, and are therefore equal to two

Similarly, the angles below E B are equal to two right angles.

Therefore all the angles at A are together equal to four right angles.

CHAPTERS III. AND IV.

TRIANGLES.

SECTION I.-The Equality of Triangles.

6. Two straight lines drawn from any point in the perpendicular bisector of a straight line to the extremities of the straight line are equal to one another.

Let A be a point in the line A D which bisects. BC and is at right angles to BC at the point D; then shall C A = BA. The triangles ADB and ADC have BD, DA, and the included angle BDA equal respectively to CD, DA and the included angle CDA.

B

D

Consequently the triangles are equal in all respects, and ABAC.

7. If from a point within a triangle two straight lines be drawn to the extremities of the base, the angle they contain shall be greater than the vertical angle of the triangle.

Let D be a point within ▲ A B C, and let D be

E

angle of A Deb,

joined to the points A and B ; then shall ADB be greater than LAC B.

Continue A D to meet B C in E.

Since ADB is the exterior

Therefore A D B is greater than ▲ DE B. Since DEB is the exterior angle of ▲ ACE, Therefore DE B is greater than ACB; Consequently ▲ ADB is greater than AC B.

8. Only one perpendicular can be drawn from a point to a straight line.

Let C be the point and A B the straight line. Let CD be perpendicular to A B, then there cannot be another perpendicular from C to A B. If possible let CE be perpendicular to A B.

Λ E

D

F

B

Then CED and CDB will be equal, because they are right angles; that is to say, the exterior angle of a triangle will be equal to the interior and opposite angle, which is impossible.

Therefore CE cannot be perpendicular to AB, and, consequently, C D is the only perpendicular from C to A B.