From this Proposition we draw the following important inferences.

1. If two triangles have two angles of the one equal to two angles of the other, each to each, then the third angle of the one is equal to the third angle of the other.

2. In any right-angled triangle the two acute angles are complementary.

3. In a right-angled isosceles triangle each of the equal angles is half a right angle.

4. If one angle of a triangle is equal to the sum of the other two, the triangle is right-angled.

5. The sum of the angles of any quadrilateral figure is equal to four right angles.

6. Each angle of an equilateral triangle is two-thirds of a right angle.

EXERCISES ON PROPOSITION 32.

1. Prove that the three angles of a triangle are together equal to two right angles,

(i) by drawing through the vertex a straight line parallel to the base;

(ii) by joining the vertex to any point in the base.

2. If the base of any triangle is produced both ways, shew that the sum of the two exterior angles diminished by the vertical angle is equal to two right angles.

3. If two straight lines are perpendicular to two other straight lines, each to each, the acute angle between the first pair is equal to the acute angle between the second pair.

4. Every right-angled triangle is divided into two isosceles triangles by a straight line drawn from the right angle to the middle point of the hypotenuse.

Hence the joining line is equal to half the hypotenuse.

5. Draw a straight line at right angles to a given finite straight line from one of its extremities, without producing the given straight line.

[Let AB be the given straight line. On AB describe any isosceles triangle ACB. Produce BC to D, making CD equal to BC. Join AD. Then shall AD be perpendicular to AB.]

6. Trisect a right angle.

7. The angle contained by the bisectors of the angles at the base of an isosceles triangle is equal to an exterior angle formed by producing the base.

8. The angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles.

The following theorems were added as corollaries to Proposition 32 by Robert Simson, who edited Euclid's text in 1756.

COROLLARY 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Let ABCDE be any rectilineal figure.

Take F, any point within it,

and join F to each of the angular points of the figure.

Then the figure is divided into as many triangles as it has sides.

I. 32.

And the three angles of each triangle are together equal to two right angles. Hence all the angles of all the triangles are together equal to twice as many right angles as the figure has sides. But all the angles of all the triangles make up all the interior angles of the figure, together with the angles at F, which are equal to four right angles. I. 15, Cor. Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Q.E.D.

COROLLARY 2. If the sides of a rectilineal figure, which has no re-entrant angle, are produced in order, then all the exterior angles so formed are together equal to four right angles.

I. 13.

For at each angular point of the figure, the interior angle and the exterior angle are together equal to two right angles. Therefore all the interior angles, with all the exterior angles, are together equal to twice as many right angles as the figure has sides.

But all the interior angles, with four right angles, are together equal to twice as many right angles as the figure has sides.

I. 32, Cor. 1. Therefore all the interior angles, with all the exterior angles, are together equal to all the interior angles, with four right angles.

Therefore the exterior angles are together equal to four right angles.

EXERCISES ON SIMSON'S COROLLARIES.

Q.E.D.

[A polygon is said to be regular when it has all its sides and all its angles equal.]

of

1. Express in terms of a right angle the magnitude of each angle (i) a regular hexagon, (ii) a regular octagon.

2. If one side of a regular hexagon is produced, shew that the exterior angle is equal to the angle of an equilateral triangle.

3. Prove Simson's first Corollary by joining one vertex of the rectilineal figure to each of the other vertices.

4. Find the magnitude of each angle of a regular polygon of n sides.

5. If the alternate sides of any polygon be produced to meet, the sum of the included angles, together with eight right angles, will be equal to twice as many right angles as the figure has sides.

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel.

Let AB and CD be equal and parallel straight lines; and let them be joined towards the same parts by the straight lines AC and BD.

Then shall AC and BD be equal and parallel.

Proof. Then because AB and CD are parallel, and BC meets them,

therefore the angle ABC is equal to the alternate angle

DCB.

Because

Now in the triangles ABC, DCB,
AB is equal to DC,

and BC is common to both;

I. 29.

Hyp.

also the angle ABC is equal to the angle

DCB;

Proved. therefore the triangle ABC is equal to the triangle DCB in all respects;

so that the base AC is equal to the base DB,
and the angle ACB equal to the angle DBC.

But these are alternate angles.
Therefore AC and BD are parallel :

and it has been shewn that they are also equal.

I. 4.

I. 27.

Q.E.D.

DEFINITION. A Parallelogram is a four-sided figure

whose opposite sides are parallel.

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

B

Let ACDB be a parallelogram, of which BC is a diagonal. Then shall the opposite sides and angles of the figure be equal to one another; and the diagonal BC shall bisect it

Proof. Because AB and CD are parallel, and BC meets them, therefore the angle ABC is equal to the alternate angle

DCB;

I. 29.

Again, because AC and BD are parallel, and BC meets them, therefore the angle ACB is equal to the alternate angle

DBC.

I. 29.

Hence in the triangles ABC, DCB, the angle ABC is equal to the angle DCB, Because and the angle ACB is equal to the angle DBC; also the side BC is common to both;

therefore the triangle ABC is equal to the triangle DCB in all respects;

I. 26.

so that AB is equal to DC, and AC to DB; and the angle BAC is equal to the angle CDB. Also, because the angle ABC is equal to the angle DCB, and the angle CBD equal to the angle BCA, therefore the whole angle ABD is equal to the whole angle

DCA.

And the triangles ABC, DCB having been proved equal in all respects are equal in area.

Therefore the diagonal BC bisects the parallelogram ACDB.

Q.E.D.