In any right-angled triang.e, the square described on the hy pothenuse is equivalent to the sum of the squares on the other troe sides.

Let ABC be a right-angled triangle, having the right angle BAC; the square described upon the side BC is equivalent to the sum of the squares upon BA, AC.

On BC describe the square BCED, and on BA, AC the squares BG, CH; and through A draw AL parallel to BD, and join AD, FC.

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Then, because each of the angles BAC, BAG is a right angle, CA is in the same straight line with AG (Prop. III., B. I.). For the same reason, BA and AH are in the same straight line.

The angle ABD is composed of the angle ABC and the right angle CBD. The angle FBC is composed of the same angle ABC and the right angle ABF; therefore the whole angle ABD is equal to the angle FBC. But AB is equal to BF, being sides of the same square; and BD is equal to BC for the same reason; therefore the triangles ABD, FBC have two sides and the included angle equal; they are therefore equal (Prop. VI., B. I.).

But the rectangle BDLK is double of the triangle ABD, because they have the same base, BD, and the same altitude, BK (Prop. II., Cor. 1); and the square AF is double of the triangle FBC, for they have the same base, BF, and the same altitude, AB. Now the doubles of equals are equal to one another (Axiom 6, B. I.); therefore the rectangle BDLK is equivalent to the square AF.

In the same manner, it may be demonstrated that the rectangle CELK is equivalent to the square AI; therefore the whole square BCED, described on the hypothenuse, is equivalent to the two squares ABFG, ACIH, described on the two other sides; that is,

Cor. 1. The square of one of the sides of a right-angled

triangle, is equivalent to the square of the hypothenuse, diminished by the square of the other side; that is,

AB' BC-AC'.

Cor. 2. The square BCED, and the rectangle BKLD, having the same altitude, are to each other as their bases BC, BK (Prop. III.). But the rectangle BKLD is equivalent to the square AF; therefore,

BC2: AB2:: BC: BK.

In the same manner,

BC2: AC2:: BC: KC.

Therefore (Prop. IV., Cor., B. II.),

AB2: AC2:: BK : KC.

That is, in any right-angled triangle, if a line be drawn from the right angle perpendicular to the hypothenuse, the squares of the two sides are proportional to the adjacent segments of the hypothenuse; also, the square of the hypothenuse is to the square of either of the sides, as the hypothenuse is to the segment adjacent to that side.

Cor. 3. Let ABCD be a square, and AC its diagonal; the triangle ABC being right-angled and isosceles, we have

AC AB2+BC2=2AB2;

therefore the square described on the diagonal of a square, is double of the square described on a side. If we extract the square root of each memher of this equation, we shall have

B

AC AB/2; or AC: AB:: /2: 1.

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PROPOSITION XII. THEOREM.

In any triangle, the square of a side opposite an acute angle, is less than the squares of the base and of the other side, by twice the rectangle contained by the base, and the distance from the acute angle to the foot of the perpendicular let fall from the opposite angle.

Let ABC be any triangle, and the angle at C one of its acute angles, and upon BC let fall the perpendicular AD from the opposite angle; then will

AB-BC+AC-2BC XCD. First. When the perpendicular falls within the triangle ABC, we have BD=BC - CD, and therefore BD'=BC+CD2—2BC ×CD (Prop. IX.). To each of these equals add AD; then BD2+AD2=BC2+CD2+AD3— 2BC XCD. But it the right-angled triangle B

A

D

ABD, BD2+AD'=AB'; and in the triangle ADC, CD'+ AD' AC' (Prop. XI.); therefore

AB'=BC'+AC-2BC XCD.

Secondly. When the perpendicular falls A without the triangle ABC, we have _BD= CD-BC, and therefore BD' CD'+BC22CD × BC (Prop. IX.). To each of these equals add ÀD'; then BD+AD' CD'+AD' +BC-2CD x BC. But BD2+AD2=AB' ; and CD'+AD' AC; therefore

AB'=BC'+AC-2BC x CD.

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Scholium. When the perpendicular AD falls upon AB, this proposition reduces to the same as Prop. XI., Cor. 1.

PROPOSITION XIII. THEOREM.

In obtuse-angled triangles, the square of the side opposite tre obtuse angle, is greater than the squares of the base and the <ther side, by twice the rectangle contained by the base, and the distance from the obtuse angle to the foot of the perpendicular let fall from the opposite angle on the base produced.

A

Let ABC be an obtuse-angled triangle, having the obtuse angle ABC, and from the point A let AD be drawn perpendicular to BC produced; the square of AC is greater than the squares of AB, BC by twice the rectangle BC × BD. For CD is equal to BC+BD; therefore CD1 =BC2+BD2+2BC × BD (Prop. VIII.). To each of these equals add AD'; then CD'+ AD2= BC2+BD2+AD2+2BCX BD. But AC' is equal to CD'+AD2 (Prop. XI.), and AB' is equal to BD2+AD'; therefore AC2=BC'+ AB2+2BC x BD. Therefore, in obtuse-angled triangles, &c.

D B

Scholium. The right-angled triangle is the only one in which the sum of the squares of two sides is equivalent to the square on the third side; for, if the angle contained by the two sides is acute, the sum of their squares is greater than the square of the opposite side; if obtuse, it is less.

In any triangle, if a straight line is drawn from the vertex to the middle of the base, the sum of the squares of the other two sides is equivalent to twice the square of the bisecting line, together with twice the square of half the base.

Let ABC be a triangle having a line AD drawn from the

middle of the base to the opposite angle; the squares of BA and AC are together double of the squares of AD and BP From A draw AE perpendicular to BC;

then, in the triangle ABD, by Prop. XIII., AB' AD+DB'+2DBX DE;

=

and, in the triangle ADC, by Prop. XII., ACAD2+DC2 —2DC ×DE.

Hence, by adding these equals, and observing that BD=DC, and therefore BD2= B DC, and DBX DE=DCX DE, we obtain

AB'+AC2=2AD2+2DB'.

Therefore, in any triangle, &c.

A

DE

PROPOSITION XV. THEOREM.

In every parallelogram the squares of the sides are together equivalent to the squares of the diagonals.

D

Let ABCD be a parallelogram, of which A the diagonals are AC and BD; the sum of the squares of AC and BD is equivalent to the sum of the squares of AB, BC, CD, DA. The diagonals AC and BD bisect each other in E (Prop. XXXII., B. I.); therefore, in the triangle ABD (Prop. XIV.),

AB'+AD'=2BE2+2AE';

and, in the triangle BDC,

CD2+BC2=2BE2+2EC2.

B

Adding these equals, and observing that AE is equal to EC, we have

AB2+BC2+CD2+AD-4BE2+4AE2.

But 4BE2=BD', and 4AE' AC' (Prop. VIII., Cor.); therefore

AB+BC+CD'+AD' BD2+AC2. Therefore, in every parallelogram, &c.

PROPOSITION XVI. THEOREM.

If a straight line be drawn parallel to the base of a triangle, it will cut the other sides proportionally; and if the sides be cut proportionally, the cutting line will be parallel to the base of the triangle.

Let DE be drawn parallel to BC, the base of the triangle ABC; then will AD: DB:: AE: EC.

Join BE and DC; then the triangle BDE is equivalent to the triangle DEC, because they have the same base, DE, and the same altitude, since their vertices B and C are in a line parallel to the base (Prop. II., Cor. 2).

The triangles ADE, BDE, whose common vertex is E, having the same altitude, are to each other as their bases AD, DB (Prop. VI., B Cor. 1); hence

ADE: BDE:: AD: DB.

The triangles ADE, DEC, whose common vertex is D, having the same altitude, are to each other as their bases AE, EC; therefore

ADE: DEC:: AE: EC.

But, since the triangle BDE is equivalent to the triangle DEC, therefore (Prop. IV., B. II.),

AD: DB:: AE : EC.

Conversely, let DE cut the sides AB, AC, so that AD: DB ::AE: EC; then DE will be parallel to BC.

For AD: DB:: ADE: BDÊ (Prop. VI., Cor. 1); and AE :EC:: ADE: DEC; therefore (Prop. IV., B. II.), ADE: BDE:: ADE: DEC; that is, the triangles BDE, DEC. have the same ratio to the triangle ADE; consequently, the triangles BDE, DEC are equivalent, and having the same base DE, their altitudes are equal (Prop. VI., Cor. 2), that is, they are between the same parallels. Therefore, if a straight line, &c.

A

Cor. 1. Since, by this proposition, AD: DB:: AE: EC; by composition, AĎ+DB: AD :: AE+EC: AE (Prop. VI., B. II.), or AB : AD:: AC: AE; also, AB: BD:: AC: EC. Cor. 2. If two lines be drawn parallel to the base of a triangle, they will divide the other sides proportionally. For, because FG is drawn parallel to BC, by the preceding proposition, AF: FB:: AG: GC. Also, by the last corollary, because DE is parallel to FG, AF : DF .: AG: EG. Therefore DF: FB:: EG: GC (Prop. IV., Cor., B. II.). Also, AD: DF:: AE: EG.

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Cor. 3. If any number of lines be drawn parallel to the base of a triangle, the sides will be cut proportionally.

PROPOSITION XVII. THEOREM.

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