=

ADC, since BD = DC, hence also the triangle BOD = COD (Ax. 3).

Again, the triangles ABD and CBE are equal, being each half of the whole triangle ABC, and the part EBDO is common, therefore the remainder the triangle AEO = COD, which again is equal to OBD. The six triangles into which the whole triangle is divided are therefore all equal; hence the triangle ABO, which is made up of two of these equal triangles, is double of DBO; but these triangles have the same altitude, and therefore the base AO is double of OD. In the same manner it can be proved that BO is double of OF, and that CO is double of OE.

Lastly, in the triangle AOB the base AB is bisected in E, and EO drawn to the vertex; therefore AO2 + OB2 = 2AE2 + 2EO2 (II. A), hence also 2AO2 + 20B2 = 4AE2 + 4E02, but 4AE2 = AB2, and 4EO2 = OC2 since OC was proved double of OE; therefore changing the sides and substituting these values,

=

AB2 + OC2

=

2AO2 + 20B2, take OC2 from each, and there remains AB2 =2A02 + 20B2 – OC2;

in the same manner it can be proved that

BC2 = 2BO2 + 20C2 – AO2, and that AC2 = 20C2 + 2A02 — BO2,

and adding these three,

AB2+BC2 + AC2= 3AO2 + 3B02 + 30C2.

EXERCISE XII.-THEOREM.

If the three sides of a triangle be 4, 8, and 10, find by proposition C, Scholium, whether the triangle is acute or obtuse angled. Take 4 for the base, and 8 and 10 for the two sides, then the triangle will be obtuse or acute angled according as the perpendicular from the vertex falls on the base produced or within the triangle, that is, according as the distance of the foot of the perpendicular from the middle of the base is greater than or less than half the base, and it is evident that if it were equal to half the base, the triangle would be right angled.

Now half the base is 2, and the distance of the foot of the perpendicular from the middle of the base is by (II. C, Scholium),

than half the base, the triangle is obtuse angled.

EXERCISE XIII.-THEOREM.

If perpendiculars be drawn from the three angles of a triangle to the opposite sides, prove that the sum of the squares on the alternate segments of the sides are equal to one another.

Let ABC be a triangle, and AD, BF, and CE perpendiculars from the angles to the opposite sides, to prove that the squares on BD, CF, and AE are together equal to the squares on CD, AF, and BE. In the triangle ABC, since AD is drawn from the vertex perpendicular to the base, AB2 — AC2 = BD2 - DC2 (II. C); to each add AC2 + DC2; therefore AB + DC2 = AC2 + BD2; that is, the squares on the one side and the opposite segment of the base are equal to the squares on the other side and its opposite segment; hence considering A, B, and C successively as the vertices, the three following equalities are obtained: AB2 + DC2 = AC2 + BD2, CB2 + AF2

and

AB2 + CF2,

AC2 + EB2 = BC2 + AE2.

B

F

D C

Adding these three, and omitting from both sides the common squares AB2 + BC2 + CA2, we have

DC2+AF2 + EB2 BD2 + CF2 + AE2; that is, the sum of the squares on the alternate segments are equal to one another.

EXERCISE XIV.-THEOREM.

If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, twice the rectangle contained by that side and the segment of it, between the base and perpendicular, is equal to the square on the base.

Let ABC be an isosceles triangle, and BD a perpendicular from B on AC; it is required to prove that BC2= 2AC CD.

=

Since ABD is a right-angled triangle BD2: AB2 AD2 (I. 47 Cor.), and AB being equal to AC, BD2 = AC2- AD2, to each add DC2; then BD2 + DC2 = AC2 + DC2 — AD2 = AD2 +2AD DC + 2DC2 — AD2 (II. 4); hence (I. 47) BC2 = 2AD · DC + 2DC2, but AD DC DC2 AC CD (II. 3), and the doubles of these are equal; wherefore BC2 = 2AC CD.

=

EXERCISE XV.-THEOREM.

B

C

The square on the base of an isosceles triangle whose vertical angle is a right angle, is equal to four times the area of the triangle. In every right-angled triangle the area = half the product of the sides containing the right angle; therefore the product of the sides is equal to twice the area; but the sides being equal, their product is the square on one of them, and hence the square on one of the equal sides is equal to twice the area; but the vertical angle being a right angle, the square on the base is equal to the sum of the squares on the sides. Therefore the square on the base is equal to four times the area of the triangle.

C

EXERCISE XVI.-THEOREM.

If in the figure to (I. 47) the angular points of the squares be joined, thus forming a six-sided figure, prove that the sum of the squares on the six sides is equal to eight times the square on the hypotenuse of the original triangle.

In the figure DE and GH are each equal to BC, hence the squares on DE and GH are together equal to twice the square on BC; also the squares on FG and HK are equal to the squares on AB and AC, and therefore equal (I. 47) to the square on BC. Therefore DE2+ GH2 + FG2 + HK2 = 3BC2. Also

and

F

B

H

M

K

LE

DF2 = DB2 + BF2 + 2DB · BN (II. 12),
in the triangle DBF,

EK2 = EC2 + CK2 + 2EC · CM (II. 12),
in the triangle ECK.

Adding these two, and observing that DB and EC are each equal
to BC; that the squares on FB and CK are equal to the squares
on AB and AC, and hence together equal to BC2; while BN and
CM were proved equal to BO and CO (Ex. 37, First Book),
gives DF2+ EK2 = 3BC2 + 2CB BO+2BC CO;
but CB BO+BC • CO = BC2 (II. 2), and the doubles of these
are equal, whence DF2 + EK2 = 5BC2; adding this result to the
former, we obtain DE2 + DF2 + FG2 + GH2 + HK2 + KE2
3BC2+5BC2 = 8BC2.

=

EXERCISE XVII.-THEOREM.

The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of its diagonals, and four times the square of the line joining their middle points.

Let ABCD be any quadrilateral, AC, BD its diagonals, and EF the line joining their middle points, then AD2 + DC2 + CB2 + BA2 AC2+DB2 + 4EF2.

For draw DE and EB. Then, because the base AC of the triangle ACD is bisected in E, therefore

(II. A)

AD2+DC2 = 2AE2 + 2DE2;

and since AC is also the base of the triangle ABC, AB2 + BC2 = 2AE2 + 2BE2. Hence, adding equals to equals,

AD2+DC2 + AB2 + BC2 = 4AE2 + 2DE2

+2EB2.

D

But (II. A) since the base BD of the triangle DEB, is bisected in

F, 2DE2 + 2EB2 = 4DF2 + 4EF2; consequently the sum of the squares of the sides is equal to 4AE2 + 4DF2+4EF2. But (II. 8, Cor. 2) 4AE2 = AC2, 4DF2 = DB2; and hence

ÁD2+DC2 + AB2 + BC2 = AC2 + DB2 + 4EF2.

EXERCISE XVIII.-THEOREM.

The sum of the squares of two opposite sides of a quadrilateral, together with four times the square of the line joining their middle points, is equal to the sum of the squares of the other two sides and of the diagonals.

Let ABDC be any quadrilateral, EF the line joining the middle points of the sides AB, CD, and AD, BC its diagonals, it is required to prove that AB2 + CD2 + 4EF2 = AC2 + DB2 + AD2 + BC2.

For join CF and FD; then, because the base AB of the triangles ABC, ABD is bisected in F (II. A),

AC2 + CB22AF2 + 2CF2,

and BD2+ AD2 · 2AF2+2FD2; consequently, adding equals to equals, AC2 + BD2 + AD2 + BC2 = 4AF2 + 2CF2 + 2FD2.

But (II. A) 2CF2 + 2FD2 = 4CE2 + A 4EF2; wherefore the sum of the squares

of AC, BD, AD, and BC is 4AF2 + 4CE2 + 4EF2. Again (II. 8, Cor. 2), 4AF2 = AB2, and 4CE2 AC2 + BD2 + AD2 + BC2 = AB2 + CD2 + 4EF2.

E

D

B

CD2; hence

EXERCISE XIX.-THEOREM.

The squares of the sum and of the difference of two lines are together double the squares of these lines.

Let AB, BC be two lines; produce AB to D, making BD = BC; then AD is the sum, and AC the dif

ference of AB and BC, and it is required to A

prove that AD2 + AC2 is

2AB2 + 2BC2.

C B

D

For (II. 4) AD2 = AB2 + BD2 + 2AB · BD = AB2 + BC2 + 2AB BC; for BD = BC.

Also (II. 7)

AC2+2AB BC= AB2 + BC2.

And, adding equals to equals,

AD2 + AC2 + 2AB · BC = 2AB2 + 2BC2 + 2AB · BC. Taking away the quantity 2AB BC from these equals, there remains, AD2 + AC2 = 2AB2 + 2BC2.

EXERCISE XX.-THEOREM.

If a line be cut in medial section, the line composed of it and its greater segment is similarly divided.

Let the line AC be cut in medial section at the point B; then, if CA be produced to D, till AD = AB, the line CD is also cut in medial section in A.

The sum of the squares of the diagonals of a quadrilateral, is equal to twice the sum of the squares of the lines joining the middle points of the opposite sides.

Let ABCD be any quadrilateral, and EG, FH lines joining the middle points of the opposite sides, and AC, DB its diagonals; then AC2 + DB2 = 2EG2 +- 2FH2.

For draw EH, HG, GF, and EF; then, by (Ex. I. 50) EHGF is a parallelogram; and DB is double of GH (Ex. 49, First Book), and AC double of EH; consequently (II. 8, Cor. 2)

AC2 + DB2 = 4EH2 + 4HG2.

But (I. 34) 4EH2 2EH2 + 2FG2;

and

4HG2

=

2HG2+2EF2.

Also (II. B) EH2 + HG2 + GF2 + EF2 = EG2
+FH2; and the double of the squares of the
sides EH, HG, GF, FE are = 2EG2 + 2FH2,
and also to 4EH2 + 4HG2;
hence

AC2+ DB2 2EG2 + 2FH2.

D

H

C

EXERCISE XXII.-THEOREM.

If the vertical angle of a triangle be four-thirds of a right angle, the square on the base is equal to the sum of the squares on the sides, together with the rectangle contained by the sides; and if the vertical angle be two-thirds of a right angle, the square on the base is equal to the sum of the squares on the two sides, diminished by the rectangle contained by the sides.

=

Let BAC be a triangle, having its vertical angle four-thirds of a right angle; it is required to prove that CB2 = CA2 + AB2 + AB AC. Draw BE perpendicular to CA produced; then since BAC, the exterior angle of the triangle ABE, is four-thirds of a right angle, and the angle at E is a right angle, the angle ABE is a third of a right angle (I. 32); hence AB = 2AE (Ex. 17, Cor., First Book).