THEOREM XIII.

If from the right angle of a right-angled triangle a line is drawn perpendicular to the hypotenuse:

I. The two partial triangles thus formed will be similar to the whole triangle, and consequently similar to each other.

II. The perpendicular will be a mean proportional between the two segments of the hypotenuse.

A

III. Either side containing the right angle will be a mean proportional between its adjacent segment and the hypotenuse. First. The two triangles ABC, ABD are similar, since they have the common angle at B, and the angle BAC of the one equal to the angle BDA of the other, each being a right angle. For the same reason, ACB is similar to ACD, consequently these triangles are similar to each other.

B

D

Secondly. Comparing the two partial triangles ABD, ACD, we have

BD: AD::AD: DC.

Thirdly. Comparing the total triangle ABC with its partial triangle ABD, observing that in these triangles BC and AB are homologous, being the hypotenuses, that AB and BD are also homologous, being opposite the equal angles BCA, BAD, we shall obtain the proportion,

BC:AB::AB: BD.

The comparison of the triangles ABC, ACD will give, in like manner,

BC: AC:: AC: DC.

THEOREM XIV.

In any right-angled triangle, the square, or second power, of the numerical value of the hypotenuse, is equal to the sum of the squares of the numerical values of the other two sides.

For, taking the product of the means and extremes of the last two proportions of the preceding Theorem, we have

A

B

D

AB2=BC × BD; AC2=BC × DC;

from which, by adding member to member,

that is,

AB2 + AC2=BC × (BD+DC) = BC × BC;

=

Cor. When AB AC, we have BC2=2AB2; that is, in any right-angled isosceles triangle, the square of the hypotenuse is double the square of one of the sides. Consequently, the second power of the numerical value of the diagonal of a square is double that of the side. And the diagonal and side must be to each other in the ratio of ✓2 to 1; that is, they are incommensurable, as has already been shown (B. II., P. XIV.).

Scholium I. From the two equations,

AB2 = BCX BD, AC2 = BC × DC,

we deduce the proportion,

AB: AC2:: BC × BD : BC × DC, or,:: BD : DC.

In a similar manner, by combining the identical equation BC2=BC2, with the same two equations above, we have

BC2: AB2:: BC × BC: BC × BD, or, :: BC: BD;
BC2: AC2:: BC BC: BC × DC, or, :: BC: DC.

and
From this we see that, in any right-angled triangle, the squares
or second powers of the sides containing the right angle and of
the hypotenuse, are proportional to the segments of this hypot
enuse, and the hypotenuse itself. That is to say, we have

AB2: AC2: BC2:: BD: DC: BC.

Scholium II. The segments BD, DC, formed by the perpendicular AD on the hypotenuse BC, are called the projections of the two sides on the hypotenuse.

In general, the projection of any definite straight line MN, on another indefinite straight line AB, is the distance PQ of the intersection of the perpendiculars. drawn through the extremities of the first line to the second.

M

Ꭺ

P

N

يم

R

B

If through M we draw MR parallel to PQ, we shall have MR PQ, and the right-angled triangle MNR will give

MN2 = MR2+NR2 = PQ2 + (NQ — MP)2.

That is-The square of a straight line of determinate length is equal to the square of its projection on another line, plus the square of the difference of the perpendiculars which determine this projection.

THEOREM XV.

In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of either of the sides containing the obtuse angle into the projection of the other side on the prolongation of the first.

If the triangle ABC is obtuse-angled at A, we shall have

BC2 AB2+AC2+2AB × AD.

=

For, since BDAB+ AD, if we square both members of this equation, using the algebraic formula, we shall have

D

Adding now, to each member of this, CD2, and observing that BD2+CD2 = BC2, and also that AD2 + CD2 = AC2, we shall obtain

BC2AB+ AC2+2AB × AD.

THEOREM XVI.

In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, diminished by twice the product of one of these sides, by the projection of the other on the preceding one, produced if necessary.

BC2 AB2+AC2-2AB-AD. For, BDAB-AD, or AD .- AB, according to which di

A

ДА

D

BA

B

D

agram we use; but in both cases BD is the difference of AB

and AD; consequently, if we take the squares of both members, by the well-known algebraic formula, we sha.1 have

BD2 AB2+ AD2-2AB X AD.

=

Adding to each member of this DC, and observing that BD2 + CD2 = BC2, and also that AD2+DC2 = AC2, we shall obtain

BC2 AB2+ AC2-2AB X AD.

=

=

This result will hold good even when the perpendicular CD coincides with the side CB, in which case AD AB, and the above will become BC2 = AC2 – AB2, or, AC2 = AB2+ BC2, which we know to be true, since the triangle ABC is, in this case, right-angled at B (T. XIV.).

Scholium. A triangle is right-angled, acute-angled, or obtuseangled, according as the square of the longest side is equal, less, or greater than the sum of the squares of the other two sides.

If a triangle has 3, 4, and 5 for the numerical values of its sides, it will be right-angled, since

52 = 32 +42.

One having 4, 5, and 6 for its sides, is acute-angled, since

62 <42+52.

One having 2, 3, and 5 for its sides, is obtuse-angled, since

52>22+32.

Thus we are furnished with a second very simple method for determining the kind of triangle, when its sides are given (B. I., T. XXXV., S.).

THEOREM XVII.

In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side, increased by twice the square of the line drawn from the middle of this third side to the opposite angle.

If CD is drawn bisecting AB, we shall

have

AC2+BC22AD2+2CD2.

For, the two triangles ADC, CDB, the

C

one obtuse-angled and the other acute-angled at D, give, by the

two preceding Theorems,

ACAD2+CD2+2AD × DE,

BC2BD2+CD2- 2DB+DE;

adding these, and observing that AD = DB, we have

AC2+BC22AD2 + 2CD2.

Cor. The sum of the squares of the four sides of any parallelogram is equal to the sum of the squares of its diagonals. This proposition is easily deduced from the above.

DETERMINATION OF AREAS.

THEOREM XVIII

Two parallelograms having equal bases and the same altitude, are equivalent.

We will suppose the one LF D I E' CF figure placed upon the other,

so that their inferior bases, which are equal, may coincide. Let ABCD be the

L

B

E L'

first parallelogram, and ABEF or ABE'F' the second. Since these parallelograms have the same altitude, their superior bases will be situated in the same indefinite line LL', parallel to the lower base.

If now, we confine our attention to the two parallelograms ABCD, ABEF, we see that the triangles BCE, ADF are equal, having equal angles CBE and DAF (B. I., T. III.) included between equal sides, namely, BE=AF, BC=AD (B. I., T. XXXI.). If, now, from the quadrilateral ABED we subtract the triangle BEC, we shall have left the parallelogram ABCD; but if from the same quadrilateral we subtract the equal triangle AFD, we shall have left the parallelogram ABEF. Hence the two parallelograms are equivalent.

We may in the same way prove that the parallelograms ABCD, ABE'F' are equivalent.