293. Theorem VIII. If two polygons are similar, they can be divided into the same number of triangles, similar each to each, and similarly placed.

294. Theorem IX. If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar.

295. Theorem X. The areas of similar polygons have the same ratio as the squares of any pair of corresponding sides.

426. If similar polygons are constructed on the sides of a right triangle as corresponding sides, the polygon on the hypotenuse equals the sum of the other two polygons.

427. If each side of one polygon is double the corresponding side of a second similar polygon, what relation have the areas?

428. If the area of one polygon is 36 times the area of a similar polygon, what relation have the sides?

NOTE. Area ratios in similar figures are always the squares of line ratios; line ratios are the square roots of area ratios.

296. Theorem XI. In a right triangle, the altitude to the hypotenuse divides the triangle into two triangles similar to each other, and to the whole triangle.

297. COR. 1. (1) The altitude to the hypotenuse is the mean proportional between the sects of the hypotenuse.

(2) Either leg of the triangle is the mean proportional between the hypotenuse and its own projection on the hypotenuse.

429. In a right triangle of legs 5 and 12, find the projections of 5 and of 12, on the hypotenuse; find the altitude to the hypotenuse.

430. In a right triangle of legs a, b, hypotenuse c, find the projections of a and of b on c; find the altitude to c.

NOTE. The results of exercise 430 are formulas that hold for all right triangles.

431. Tangents from a point to a circle of radius 6 are of length 8. Find the chord of contact.

432. In a right triangle, the sects of the hypotenuse made by the altitude are 4 and 5. Find the other sides and the altitude.

433. The squares of two chords drawn from a point on a circle have the same ratio as their projections on the diameter from that point.

434. The half chord perpendicular to a diameter is the mean between the sects of the diameter.

298. Theorem XII. A line that bisects an angle, interior or exterior, of a triangle, divides the opposite side, internally or externally, into sects proportional to the other sides of the triangle. Note that the two lines divide harmonically.

435. If a line drawn from the vertex of an angle of a triangle divides the opposite side, internally or externally, in the ratio of the other two sides, the line bisects the angle, interior or exterior, from whose vertex it is drawn.

436. In a triangle of sides 6, 7, 8, find the sects of 7 made by the bisector of the opposite angle; of the exterior angle at the opposite vertex.

437. In a triangle of sides a, b, c, find the sects of c made by the bisector of the opposite angle; of the exterior angle at the opposite vertex.

NOTE. The results of 437 are formulas that hold for all triangles.

438. If two lines from the vertex of the right angle of a right triangle make equal angles with one of the legs, they cut the hypotenuse harmonically.

299. Theorem XIII. If a pencil of lines cuts a circumference, the lines are cut proportionally, so that the product of the two sects from the vertex to the circumference on one line is the same as that product on any other line.

300. COR. 1. A tangent from the vertex of a pencil to a circumference is the mean proportional between the two sects, from the vertex to the circumference, of any other line of the pencil that is cut by the circumference.

439. If two points are taken on each line of a pencil, so that the product of the two sects from the vertex is the same for all the lines of the pencil, the two points being on opposite sides of the vertex, the four points on any two lines are concyclic.

440. In the figure of 439, if the two points on each line are on the same side of the vertex, the four points on any two lines are concyclic.

441. Find the locus of the point P on a secant to a given circle which cuts the circle in changing points, A and B, so that PA × PB is constant.

442. If two intersecting lines are cut, one by one point, the other by two points on the same side of the vertex, so that the sect cut off by the one point is the mean proportional between the sects from the vertex on the other line, a circle through the three points would be tangent to the line cut by the one point.

443. Construct a circle through two given points tangent to a given line.

444. The product of two sides of a triangle equals the square of the bisector of the included angle, plus the product of the sects into which the bisector divides the opposite side.

SUGGESTION. Inscribe the triangle.

445. If three circles intersect, their common chords are concurrent.

446. If two circles intersect, tangents from any point in their common chord extended, to the two circles, are equal.

SECTION IV. CONSTRUCTIONS

301. CONST. I. To divide a given sect internally, and externally, in a given ratio (or, harmonically).

NOTE. A given ratio is always represented by two given sects, the given ratio being that of those sects.

302. COR. 1. To divide a given sect into parts proportional to any number of given sects.

447. Given a sect, and one point of harmonic division, find the other.

448. Given a sect, cut it into parts having the ratio 1:2:5.

303. CONST. II. To find the fourth proportional to three given sects.

304. COR. 1. To find the third proportional to three given sects.

449. Draw a line through a given point so that it will cut off sects having a given ratio on the arms of a given angle.

450. Draw a line from a given point to a given line, so that it will have a given ratio to the perpendicular from that point.

451. Draw a line through a given point so that it will be cut in a given ratio by the arms of the angle.

452. Construct two sects, given their sum and their ratio.

453. Construct two sects, given their difference and their ratio.

305. CONST. III. To find the mean proportional between two given sects. Find three methods.

This construction is the foundation of all square root questions in the Geometry. Since the square of the mean

equals the product of the extremes, it follows that the mean equals the square root of the product of the extremes. It is, therefore, possible to construct the square root of any required number by making the extremes of such length that their product equals the number of which the square root is asked. This is used also in constructing figures whose areas have a certain ratio, for if the figures are similar, the line ratio is the square root of the given area ratio, and so can be found by the mean proportional.

454. Construct a sect √2 in. long, given the sect 1 in. long.

455. Explain how the sect √n in. long could be constructed for any value of n, given a sect of 1 in.

456. Find another way to construct the square root of a number.

306. Mean and Extreme Ratio. If a sect is divided so that the longer part is the mean between the whole sect and the shorter part, the sect is said to be divided in mean and extreme ratio. If AB is divided by P so that

AB AP

= then P divides AB in mean and extreme ratio. AP PB

There is an external mean and extreme division as well as an internal; the words " mean and "extreme" do not refer to the interior and exterior cases, but to the position in which the parts occur in the proportion.

It should be noticed that, as the proportion stands, there are two unknown sects used; by a proper transforming of the proportion the parts can be combined (in different. ways for the two cases) so that the given sect AB will appear more often, thus displacing the unknowns. The new form, in which the unknowns occur as seldom as possible, is the one with which to work in attempting the construction; it will be found very easy if attacked logically.