EXAMPLES XII.

1. Divide a given straight line into four equal parts. 2. Find the eighth part of a given finite straight line.

3. Prove that the following construction bisects the line AB, with A as centre and any radius greater than half AB describe a circle; with B as centre and with radius equal to that of the first circle describe another circle cutting the first in D, E; join DE cutting AB in its middle point F.

4. Divide an equilateral triangle into four equilateral triangles.

A Right Angle.

74. DEF. When a straight line standing upon another straight line makes the adjacent angles equal to each other, each of these angles is called a right angle.

ID

B

Thus, when ACB is a straight line, and the angle ACD is equal to the angle BCD, then each of the angles ACB, BCD is a right angle.

The word right has here the meaning of the word upright.

75. The straight line which stands upon the other is said to be perpendicular to it.

Thus in the above figure, CD is perpendicular to ACB.

Also the lines ACB and DC are said to be at right angles to each other.

Proposition 11.

76. To draw a straight line at right angles to a given straight line from a given point in the same.

Let ACB be the given straight line, and C the given point in it;

it is required to draw from C a straight line at right angles to AB.

EB

Construction. In AC take a point D.
From CB cut off CE equal to CD.

Upon DE describe the equilateral triangle DFC. [Prop. 1]

Join CF.

Then CF is the straight line required.

Proof. Consider the triangles DFC, EFC;

because the sides CD, CF, FD of the triangle DFC are equal respectively to the sides CE, CF, FE of the triangle EFC; therefore, by Prop. 8,

the triangles DCF, ECF are equal in all respects;
so that the angle FCD

is equal to the corresponding angle FCE.
Again, because ECD is a straight line,

and the adjacent angles FCD, FCE are equal,

therefore each of the angles FCD, FCE is a right angle.

[Def.]

Wherefore, CF has been drawn from the point C at right

angles to the line ACB.

Q.E.F.

Example. Find a series of points which are equidistant from two given points.

Let A, B be the two given points.

[Analysis. Suppose C to be a point equidistant from A and B, so that CA is equal to CB.

Join C to D the middle point of AB.

Then it can be shewn as in Prop. 11 that CD is perpendicular to AB.

Hence we have the following]

A

B

Construction. Bisect AB in D and draw DE at right angles to AB; in DE take any point C,

then C is a point equidistant from A and B.

Proof. Consider the triangles ADC, BDC; because the sides AD, DC and their included angle ADC in the triangle ADC are equal to the sides BD, DC and their included angle in the triangle BDC.

Therefore the triangles are equal in all respects [Prop. 4] so that the side AC is equal to the corresponding side BC.

Q.E.F.

77. The above example shews (i) that every point which is equidistant from the points A and B is on the line DC, (ii) that every point on the line CD is equidistant from the points A and B.

Hence the line DC is said to be the locus of the points which are equidistant from the two given points A and B.

Another example of a locus is the circle, which is the 'locus' of all points which are at a given distance from a given point.

Example. Find the points in a given unlimited straight line whose distance from a given point is equal to a given finite straight

line.

Let BC be the given unlimited straight line, DE the given finite straight line and A the given point.

Construction. With centre A and radius equal to DE

describe the circle FGH.

[If this circle does not cut the line BC, there are no points in it whose distance from A is equal to DE, for every point which is at the distance DE from A is on the circle FGH.]

If the circle FGH cuts BC in F and G,

then F, G are the points required.

Proof. For F, G are each on the circle FGH, and every point

on the circle is at a distance equal to DE from A.

Q.E.F.

EXAMPLES XIII.

1. Find the points whose distance from a given point A is equal to a given finite straight line CD and whose distance from another given point is equal to another given finite straight line EF.

2. A and B are two given points, CD is a given straight line; find the point in CD which is equidistant from A and B.

3. In a given triangle find the point which is equidistant from its angular points.

4. Find the points which are at a given distance AB from a given point C and which are equidistant from two given points E, F.

5. Upon a given base BC describe a triangle such that the perpendicular from the vertex on the base is equal to a given finite straight line EF and cuts the base in a given point D.

6. Describe an isosceles triangle when the length of the base and of the perpendicular from the vertex to the base is given.