the triangles DEF and DGF, having thus the sides DE and EF equal to DG and FG, and the side DF common to both, are (I. 2.) equal; consequently the angle EDF is equal to FDG or BAC, and the angle EFD is equal to DFG or BCA.

Cor. Hence isosceles triangles which have either side proportional to the base, are similar.

Scholium. The second Proposition of Book I. may be considered as only a particular case of this theorem.

PROP. XIII. THEOR.

Triangles are similar, if each have an equal angle, and its containing sides proportional.

In the triangles BAC and EDF, let the angle ABC be equal to DEF, and the sides which contain the one be proportional to those which contain the other, or AB: BC: DE: EF; the triangles BAC and EDF are similar.

For, from the points E and F, draw EF and FG, making the angles FEG and EFG equal to CBA and BCA.

The triangles BAC and EGF, having thus their corresponding angles equal, are similar (VI. 11.), and therefore AB: BC: EG: EF. But by hypothesis, AB: BC::

and DFE, having the side EG equal to ED, EF common to both, and the contained angle GEF equal to ABC or DEF, are equal (I. 3.), and therefore the angle EFG or BCA is equal to EFD; consequently the remaining angles BAC and EDF of the triangles ABC and DEF are equal (I. 30.), and these triangles are (VI. 11.) similar.

Scholium. The third Proposition of Book I. is merely a particular case of this general theorem.

PROP. XIV. THEOR.

Triangles are similar, which have each an equal angle, and the sides containing another angle of the same character proportional.

Let the triangles CAB and FDE have the angle ABC equal to DEF, and the sides that contain the angles at C and F proportional, or BC: AC:: EF: FD; while those angles are both of them either acute or obtuse, the triangles ABC and DEF are similar.

For, from the points E and F draw EG and FG,

and BC: CA :: EF: FG; but, by hypothesis, BC:CA:: EF: FD, and therefore EF: FG:: EF: FD, and FG is equal to FD. Whence the triangles EGF and EDF, having the angle FEG equal to FED, the side FG equal to FD, and the side EF common, and being both of the same character with CAB, are equal (I. 21.); consequently the angle GFE or ACB is equal to DFE, and therefore (VI. 11.) the triangles ABC and DEF are similar.

Scholium. This Theorem exhibits the general property of which Prop. 2. Book II. is only a particular case.

PROP. XV. THEOR.

A perpendicular let fall upon the hypotenuse of a right-angled triangle from the opposite vertex, will divide it into two triangles that are similar to the whole and to each other.

Let the triangle ABC be right-angled at B, from which the perpendicular BD falls upon the hypotenuse AC; the triangles ABD and DBC, thus formed, are similar to each other, and to the whole triangle ACB.

For the triangles ABD and ACB, having the angle BAC common, and the right angle ADB equal to ABC,

are similar (VI. 11.). Again, the triangles DBC and ACB are similar, since they have the angle BCD common, and the right an

B

gle BDC equal to ABC. The tri

D

angles ABD and DBC being,

therefore, both similar to the same triangle ABC, are evidently similar to each other (VI. 11.).

Cor. Hence the side of a right-angled triangle is a mean proportional between the hypotenuse and the adjacent segment, formed by a perpendicular let fall upon it from the opposite vertex; and the perpendicular itself is a mean proportional between those segments of the hypotenuse. For the triangles ABC and ADB being similar, AC: AB:: AB AD; and the triangles ABC and BDC being similar, AC: BC:: BC: CD; again, the triangles ADB and BDC are similar, and therefore AD: DB::DB: DC.

Scholium. This corollary affords an easy demonstration of the celebrated theorem contained in Prop. 10. Book I.

PROP. XVI. PROB.

To find the mean proportional between two gi ven straight lines.

Let it be required to find the mean proportional between the straight lines A and B.

Find C (III. 27.) the side of a square which is equivalent to the rectangle contained by A and B; C is

the mean proportional required.

A

BH

CH

For since CAB, it follows (V. 6.) that A: C:: C: B.

PROP. XVII. PROB.

To divide a straight line, whether internally or externally, so that the rectangle under its segments shall be equivalent to a given rectangle.

Let AB be the straight line which it is required to cut, so that the rectangle under its segments shall be equivalent to a given rectangle.

From the extremities of AB, erect the perpendiculars AD and BE, equal to the sides of the given rectangle, and in the same or in opposite directions, according as the line is to be cut inter

nally or externally; join

DE, on which, as a diameter, describe a circle, meeting AB or its extension in the point C: AC and CB are the segments required.

For join DC and CE. The angle DCE, being contained in a semicircle, is a right angle (III. 19.), and therefore, in both cases, the angles ACD and BCE are together equal to a right angle.

But the angles ACD and

CDA are likewise toge

D

E

B

B

ther equal to a right angle (I. 30. cor. 1.); and consequent