Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

therefore the sum of the squares on AD and DB together with twice the square on DE is equal to double the sum of the squares on AC, CD, and DE,

therefore the sum of the squares on AD and DB is double the sum of the squares on AC and CD.

Q.E.D.

Second Proof. Because AD is the sum of AC and CD, therefore the square on AD is greater than the sum of the squares

on AC and CD by twice the rectangle AC, CD;

and because BD is the difference of BC and CD,

that is, of AC and CD, since AC is equal to BC,

II. 6.

II. 7.

therefore the square on BD is less than the sum of the squares on AC and CD by twice the rectangle AC, CD; therefore the sum of the squares on AD and BD is double the sum of the squares on AC and CD.

Q.E.D.

COR. The sum of the squares on the sum and on the difference of two given lines is double the sum of the squares on those lines.

Ex. 25. The difference of the squares on the sum and difference of two given lines is equal to four times the rectangle contained by the lines.

Ex. 26. Show that Theor. 13 is a limiting case of Theor. 12.

SECTION II

PROBLEMS

PROB. 1. To construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle: it is required to construct a parallelogram equal to ABC, and having an angle equal to D.

Bisect BC at E,

I. Prob. 4.

at the point E in CE make the angle CEF equal to D, I. Prob. 5.

[blocks in formation]

Then each of the triangles AEB, AEC, is equal to the triangle FEC,

II. 2, Cor. 1.

therefore the triangle ABC is double the triangle FEC; and because the diagonal FC of the parallelogram FECG bisects

it,

I. 29.

therefore the parallelogram FECG is also double the triangle FEC;

therefore the parallelogram FECG is equal to the triangle ABC, and it has the angle FEC equal to the given angle D.

Q.E.F.

PROB. 2. To construct a parallelogram on a given base equal to a given triangle and having one of its angles equal to a given angle.

Let AB be the given base, C the given triangle, and D the given angle:

it is required to construct on AB a parallelogram equal to C, and having an angle equal to D.

[blocks in formation]

Construct a parallelogram GBEF equal to the triangle C, and

having the angle GBE equal to D,

so that BE may be in a straight line with AB.

II. Prob. I.

Through A draw AH parallel to BG to meet FG produced in H.

Join HB.

Because HA is parallel to FE, HB is not parallel to FE,

therefore HB and FE will meet if produced,

and on the side of BE,

I. Prob. 6.

since HB falls within the angle FHA,

Let them meet at the point K.

Through K draw KLM parallel to AE or HF, and meeting GB, HA produced in L and M ;

then shall AL be the parallelogram required.

I. Frob. 6.

Because AL and GE are the complements of the parallelograms AG and LE about the diagonal HK of the parallelogram

FM,

therefore AL is equal to GE,

but GE was made equal to C,

therefore AL is equal to C.

11. 4.

Again the angle ABL is equal to the vertically opposite angle GBE,

and therefore is equal to D.

I. 4.

Constr.

Hence AL is a parallelogram upon the base AB, equal to the triangle C, and having the angle ABL equal to D.

Q.E.F.

PROB. 3. To construct a parallelogram equal to a given rectilineal figure and having one of its angles equal to a given angle.

Let ABCD be a given figure of four sides, E the given angle :

it is required to construct a parallelogram equal to ABCD and having an angle equal to E.

E

B

K

M

Join AC. Construct the parallelogram FGHK equal to the triangle ABC, and having the angle FGH equal to E. II. Prob. 1. On KH construct the parallelogram KHLM equal to the triangle ACD, and having the angle KHL equal to E;

II. Prob. 2. then shall the figure FGLM be the parallelogram required.

Because each of the angles FGH, KHL is equal to E,

Constr.

therefore KHL is equal to FGH:

but FGH is supplementary to GHK,

I. 28.

therefore KHL is supplementary to GHK,

therefore GH and HL are in the same straight line.

I. 3.

Again, because FK, KM are each parallel to GHL,

and have a common point K,

Ax. 3.

therefore FKM is a straight line,

hence GL and FM are parallel.

And FG and ML, being each parallel to KH, are parallel to one another,

therefore the figure FGLM is a parallelogram.

Also FGLM is equal to ABCD,

I. 24.

since it is made up of FH, KL, which were made equal to the triangles ABC, ACD;

and the angle FGL was made equal to E;

therefore FGLM is the parallelogram required.

If the figure has five or more sides it can be divided into three or more triangles, and the same method of construction and proof may be used as for a four-sided figure.

Q.E.F.

« ΠροηγούμενηΣυνέχεια »