1. The sum of the squares on the sides of a triangle is equal to twice the square on half the base, together with twice the square on the straight line which joins the vertex to the middle point of the base.

Let ABC be a triangle; and let D be the middle point of the base AB. Draw CE perpendicular to the base

A A A

BE

meeting it at E; then E may be either in AB or in AB produced.

First, let E coincide with D; then the proposition follows immediately from I. 47.

Next, let E not coincide with D; then of the two angles ADC and BDC, one must be obtuse and one acute. Suppose the angle ADC obtuse. Then, by II. 12, the square on AC is equal to the squares on AD, DC, together with twice the rectangle AD, DE; and, by II. 13, the square on BC together with twice the rectangle BD, DE is equal to the squares on BD, DC. Therefore, by Axiom 2, the squares on AC, BC, together with twice the rectangle BD, DE are equal to the squares on AD, DB, and twice the square on DC, together with twice the rectangle AD, DE. But AD is equal to DB. Therefore the squares on AC, BC are equal to twice the squares on AD, DC.

2. If two chords intersect within a circle, the angle which they include is measured by half the sum of the intercepted arcs.

Let the chords AB and CD of a circle intersect at E; join AD.

The angle AEC is equal to the angles ADE, and DAE, by I. 32; that is, to the angles standing on the arcs AC and BD. Thus the angle AEC is equal to an angle at the circumference of the circle standing on the sum of the arcs AC and BD; and is therefore equal to an angle at the centre of the

circle standing on half the sum of these arcs.

Similarly the angle CEB is measured by half the sum of the arcs CB and AD.

3. If two chords produced intersect without a circle, the angle which they include is measured by half the difference of the intercepted arcs.

Let the chords AB and CD of a circle, produced, intersect at E; join AD.

The angle ADC is equal to the angles EAD and AED, by I. 32. Thus the angle AEC is equal to the difference of the angles ADC and BAD; that is, to an angle at the circumference of the circle standing on an arc which is the

difference of AC and BD; and is therefore equal to an angle at the centre of the circle standing on half the difference of these arcs.

4. To draw a straight line which shall touch two given circles.

Let A be the centre of the greater circle, and B the centre of the less circle. With centre A, and radius equal to the difference of the radii of the given circles, describe a circle; from B draw a straight line touching the circle

D

so described at C. circumference at D.

Join AC and produce it to meet the Draw the radius BE parallel to AD, and on the same side of AB; and join DE. Then DE shall touch both circles.

See I. 33, I. 29, and III. 16 Corollary.

Since two straight lines can be drawn from B to touch the described circle, two solutions can be obtained; and the two straight lines which are thus drawn to touch the two given circles can be shewn to meet AB, produced through B, at the same point. The construction is applicable when each of the given circles is without the other, and also when they intersect.

When each of the given circles is without the other we can obtain two other solutions. For, describe a circle with A as a centre and radius equal to the sum of the radii of the given circles; and continue as before, except that BE and AD will now be on opposite sides of AB. The two straight lines which are thus drawn to touch the two given circles can be shewn to intersect AB at the same point.

5. To describe a circle which shall pass through three given points not in the same straight line.

This is solved in Euclid IV. 5.

6. To describe a circle which shall pass through two given points on the same side of a given straight line, and touch that straight line.

Let A and B be the given points; join AB and produce it to meet the given straight line at C. Make a square equal to the rectangle CA, CB (II. 14), and on the

E

given straight line take CE equal to a side of this square. Describe a circle through A, B, E (5); this will be the circle required (III. 37).

Since E can be taken on either side of C, there are two solutions.

The construction fails if AB is parallel to the given straight line. In this case bisect AB at D, and draw DC at right angles to AB, meeting the given straight line at C. Then describe a circle through A, B, C.

7. To describe a circle which shall pass through a given point and touch two given straight lines.

Let A be the given point; produce the given straight lines to meet at B, and join AB. Through B draw a straight line, bisecting that angle included by the given straight lines within which A lies; and in this bisecting straight line take any point C. From C draw a perpendicular on one of the given straight lines, meeting it at D; with centre C, and radius CD, describe a circle, meeting AB, produced if necessary, at E. Join CE; and through A draw à straight line parallel to CE, meeting BC, produced if

necessary, at F. The circle described from the centre F, with radius FA, will touch the given straight lines.

For, draw a perpendicular from F on the straight line BD, meeting it at G. Then CE is to FA as BC is to BF, and CD is to FG as BC is to BF (VI. 4, V. 16). Therefore CE is to FA as CD is to FG (V. 11). Therefore CE is to CD as FA is to FG (V. 16). But CE is equal to CD; therefore FA is equal to FG (V. A).

If A is on the straight line BC we determine E as before; then join ED, and draw a straight line through A parallel to ED meeting BD produced if necessary at G; from G draw a straight line at right angles to BG, and the point of intersection of this straight line with BC, produced if necessary, is the required centre.

As the circle described from the centre C, with the radius CD, will meet AB at two points, there are two solutions.

If A is on one of the given straight lines, draw from A a straight line at right angles to this given straight line; the point of intersection of this straight line with either of the two straight lines which bisect the angles made by the given straight lines may be taken for the centre of the required circle.

If the two given straight lines are parallel, instead of drawing a straight line BC to bisect the angle between them, we must draw it parallel to them, and equidistant from them.