the greater angle lying opposite the greater side (Th. XIII ), which is contrary to the hypothesis that a and b are equal; .. the conclusion is false; .. the supposition which led to this conclusion is false; that is, it is false that BC and AC are unequal; .. it is true that they are equal.

Therefore, etc.

Cor.-Every equiangular triangle is equilateral.

For, if ab, then, as just proved, B C = A C;

and,

if a = c, then, for the same reason, B C =A B;

... AB AC=BC
=

Ax. 1.

THEOREM XIX.

If from a point without a straight line, a perpendicular and oblique lines be drawn to that line, then:

1. Oblique lines, which meet the given line at equal distances from the foot of the perpendicular, will be equal.

2. Of two oblique lines meeting the given line at unequal distances from the foot of the perpendicular, the one meeting it at the greater distance will be the longer.

3. The perpendicular will be shorter than any oblique line.

Let C be the given point, and A B the given line; CD the perpendicular; CE and CF oblique lines terminating at equal distances from D; CE and C G oblique lines terminating at unequal distances from D; then

Second: Let G D > ED, then will G C > E C.

For, since b is a right angle, a is acute, and if a is acute, e must be obtuse; and, since a triangle can have but one obtuse

Cor.—A perpendicular is the shortest distance from a point to a line.

For, from the same point to the same line only one perpendicular can be drawn (Th. XVI); hence all other lines drawn from this point to the given line must be oblique; but by the Theorem just proved, the perpendicular is shorter than any oblique line.

2. On the supposition that b is a right angle, why must a be acute? (See Th. XII.)

3. If a is acute, why must e be obtuse? (See Th. I.)

4. By what Theorem are the triangles C D E and CDF shown to be equal?

5. In the proof of 'First,' what is the important point?- That the triangles, C D E and C D F, are equal. In the proof of 'Second'?— That e is an obtuse angle. In the proof of 'Third'?- That b is a right angle.

THEOREM XX.

If a perpendicular be drawn to the middle point of a straight line, then:

1. Any point of the perpendicular will be equally distant from the extremities of the line.

2. Any point outside of the perpendicular will be unequally distant from the extremities of the line.

Let A B be the given line, C its middle point, EF the perpendicular, P any point in the perpendicular, and P' any point outside of it; then,

First: P is equally distant from A and B.

E

Р

B

D

For, draw AP and BP; then, since by hypothesis A C B C,

AP BP (Th. XIX); that is,

F

any point of the perpendicular is equally distant from the extremities of the given line, A and B.

Second: P' is unequally distant from A and B; that is, A P'> or < P'B.

For, if from P'a perpendicular be let fall on A B, it must terminate either at the right or left of C; for, if it terminated at C, there would be two perpendiculars erected to a given line from the same point of that line, which is impossible (Th. XVI, Ex.); and since it falls at the right or left of C, it will divide A B into two unequal parts, as A D and DB; hence, by the second part of Theorem XIX, A P'> or < P' B.

Therefore, if a perpendicular, etc.

QUERIES.

1. What previous Theorems are used in the above demonstration? 2. What are the essential points in this demonstration? (1) AC BC;

.. (2) A P = BP; (3) P' D cannot pass through C; (4) AD > DB; .'. (5) AP' > P' B.

2. Prove that, if a straight line has two of its points equally distant from the extremities of another line, it will be perpendicular to that line at its

If two right triangles have the hypotenuse and a side of the one equal to the hypotenuse and a side of the other, they will be equal in all their parts.

= DF, and the side BC = EF; then will the two triangles be

equal in all their parts.

For, let D E F be applied to ABC so that the angle E shall

coincide with the angle B (Ax. 6), E F coinciding with its equal B C, and ED taking the direction of B A. Now, D must fall to the right of A, or to the left of A, or upon A. If it falls to the right, DFA C; if it falls to the left, DF > AC (Th. XIX); both of which conclusions are contrary to the supposition,-A C = DF: hence, D falls upon A; DF coincides with AC (Ax. 7); and the two triangles are coincident.

Therefore, etc.

DEFINITIONS.

1. A polygon of four sides is called a Quadrilateral. Quadrilaterals are divided into three classes: trapeziums, trapezoids, and parallelograms.

2. A Trapezium is a quadrilateral having

no two sides parallel, as A.

3. A Trapezoid is a quadrilateral having

two of its opposite sides parallel, as B.

4. A Parallelogram is a quadrilateral having its opposite sides parallel, as C. Parallelograms are divided into rhomboids

and rectangles.

C

B

A

5. A Rhomboid is a parallelogram whose angles are all oblique, as C above.

6. A Rhombus is an equilateral rhomboid,

as D.

D

7. A Rectangle is a parallelogram whose angles are all right angles, as E.

E