the middle of the base BC to the vertex A: then will

2

AB2 + ä2 = 2BE2 + 2EA2.

Draw AD perpendicular to BC; then, from Proposition XII., we have,

Adding these equations, member to member (A. 2), recollect ing that BE is equal to EC, we have,

ÁВ2 + ÃỠ2 = 2BE2 + 2EA2;

which was to be proved.

Cor. Let ABCD be a parallelogram, and BD,

its diagonals. Then, since the diagonals

AC,

mutually bisect each other (B. I., P. B

C

whence, by addition, recollecting that AE is equal to CE, and BE to DE, we have,

AB2 + BC2 + CD2 + DA2 = 4CE2 + 4DE2

2

but, 4CE is equal to AC2, and 4DE to BD2 (P. VIII., C.): hence,

2

AB + BC + CD + DA AC2 + BD2.

That is, the sum of the squares of the sides of a parallelogram, is equal to the sum of the squares of its diagonals.

PROPOSITION XV. THEOREM.

In any triangle, a line drawn parallel to the base divides the other sides proportionally.

Let ABC be a triangle, and DE a line parallel to the base BC * then

AD : DB :: AE : CE.

Draw EB and DC.

Then, because

A

the triangles AED and DEB have their

bases in the same line AB, and their vertices at the same point E, they will have a common altitude: hence, (P. VI.,

C.)

AED : DEB :: AD : DB.

B

The triangles AED and EDC, have their bases in the same line AC, and their vertices at the same point D; they have, therefore, a common altitude; hence,

AED : ED C :: AE : EC.

But the triangles DEB and EDC have a common base DE, and their vertices in the line BC, parallel to DE; they are, therefore, equal: hence, the two preceding proportions have a couplet in each equal; and consequently, the remaining terms are proportional (B. II., P. IV.), hence,

AD : DB : :: AE : EC;

which was to be proved.

Cor. 1. We have, by composition (B. II., P. VI.),

AD + DB : AD

::

AE EC: AE;
+

or,

AB : AD :: AC : AE;

and, in like manner,

AB : DB :: AC : EC.

Cor. 2. If any number of parallels be drawn cutting two lines, they will divide the lines proportionally.

For, let O be the point where AB

and CD meet. In the triangle OEF, the line AC being parallel to the base EF, we shall have,

OE : AE :: OF : CF.

In the triangle OGH, we shall have,

OE : EG :: OF : FH;

hence (B. II., P. IV., C.),

A

E

F

If a line divides two sides of a triangle proportionally, it will be parallel to the third side.

Let ABC be a triangle, and let DE

divide AB and AC, so that

A

angles ADE and DEB will have a common altitude; and consequently, we shall have,

ADE : DEB :: AD : DB.

The triangles ADE and EDC have also a common altitude; and consequently, we shall have,

ADE : EDC :: AE : EC;

but, by hypothesis,

AD : DB :: AE : EC ;

hence (B. II., P. IV.),

ADE : DEB : : ADE : ED C.

The antecedents of this proportion being equal, the consequents will be equal; that is, the triangles DEB and EDC are equal. But these triangles have a common base DE: hence, their altitudes are equal (P. VI., C.); that is, the points B and C, of the line BC, are equally distant from DE, or DE prolonged parallel (B. I., P. XXX., C.) ;

hence, BC and DE are which was to be proved.

PROPOSITION XVII. THEOREM.

The line which bisects the vertical angle of a triangle, divides the base into segments proportional to the adjacent sides.

Let AD bisect the vertical angle A of the triangle BAC : then will the segments BD and DC be proportional to the adjacent sides BA and CA.

From C, draw CE parallel to DA, and produce it

until it meets BA prolonged, at E. Then, because

CE

and DA are parallel, the angles BAD and AEC are

equal (B. I., P. XX., C. 3); the angles DAC and ACE are

also equal (B. I., P. XX., C. 2). But, BAD and DAC are equal, by hypothesis; consequently, AEC and ACE are equal: hence, the triangle ACE is isosceles, AE being equal to AC.

B

D

E

In the triangle BEC, the line AD is parallel to the base EC: hence (P. XV.),

Triangles which are mutually equiangular, are similar.

Let the triangles ABC and DEF have the angle A equal to the angle D, the angle B to the angle E, and the angle C to the angle F: then will they be similar.

point H, of BC; the point D at some point G, of BA;