EXERCISES

1. The areas of two similar polygons are to each other as 121 : 169. What is the ratio of their perimeters? of their homologous diagonals? of the areas of two corresponding triangles formed by diagonals through homologous vertices? What is the ratio of similitude of two such triangles?

2. To construct a hexagon similar to a given hexagon and having an area twice as great.

3. To construct a polygon similar to two given similar polygons and equal to their sum (or difference). Construct a line (§ 349) whose square is equal to the sum (or the difference) of the squares of homologous sides of the two given polygons. On this line construct (§ 414) a polygon similar to one of the given polygons. Prove the construction.

NUMERICAL PROPERTIES OF TRIANGLES

417. Projection. The projection of one line upon another is the segment of the second line included between the feet of the perpendiculars let fall upon it from the extremities of the first line.

CD (in the last figure AD) is the projection of AB upon MN. The line upon which the projection is taken is called the base line.

418. Each leg of a right triangle is the projection of the hypotenuse upon that leg.

419. The projection of either leg of an isosceles triangle upon the base is equal to one half the base.

PROPOSITION XX. THEOREM

420. 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 those sides and the projection of the other upon it.

Given, in the triangle ABC, that p is the projection of the side b

upon the side a, and that the angle C is acute.

1. What is the projection of AB on BC, if AB is 10 in. long and makes with BC an angle of 30°? 45°? 60°? 90°?

2. Show that the projection of AB upon CD is unchanged if AB is moved toward or away from CD, remaining always parallel to its original position (see fig., p. 276).

3. Show that the projection of AB upon BC is equal to the product of AB and the cosine of the angle B.

4. Show that in the above proposition c2 = a2 + b2 – 2 ab cos C.

PROPOSITION XXI. THEOREM

421. In any obtuse 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 one of those sides and the projection of the other upon it.

Given, in the triangle ABC, that p is the projection of the side b

upon the side a, and that the angle BCA is obtuse.

To prove that

c2 = a2 + b2 + 2 ар.

Proof. 1. Draw the altitude h upon the side a.

Discussion. When the three sides of a triangle are given, the last two propositions make it possible to find the projection of any side upon any other side. The altitude upon any side may then be determined. Explain (see also § 354).

From Propositions XX and XXI may be derived a method of ascertaining from the lengths of the three sides of a triangle whether the triangle is acute, right, or obtuse.

In a right or an obtuse triangle the greatest side is opposite the right or the obtuse angle. Why? Hence if cis the greatest side of a triangle, and

c2 < a2 + b2, the triangle is acute (Proposition XX).
c2 = a2 + b2, the triangle is right.

If

If

c2 > a2 + b2, the triangle is obtuse (Proposition XXI).

In the above proposition show that c2 = a2 + b2 + 2 ab cos (180° — C).

PROPOSITION XXII. THEOREM

422. In any triangle the sum of the squares of two sides is equal to twice the square of half the third side, increased by twice the square of the median on that side.

Given the triangle ABC in which m is the median to the side c.

Proof. 1. Draw CE perpendicular to AB, and suppose that E falls between A and D. Let DE, the projection of CD upon AB, be denoted by p.

2. Then in ABCD, a2 = (2)2 + m2 + 2() p.

0-2 0-2

Why?

Why?

Ax. 2

423. COROLLARY. In any triangle the difference of the squares of two sides is equal to twice the product of the third side and the projection of the median on that side.

In the above demonstration a2 - b2 = 2 ср.

Ax. 3

424. From the result of § 422 the formula for the length of the median me upon any side of a triangle, as c, may be derived, namely, m=√2a2+ 2 b2 - c2.

PROPOSITION XXIII. THEOREM

425. In any triangle the product of two sides is equal to the square of the bisector of the included angle, increased by the product of the segments of the third side made by that bisector.

Given the triangle ABC, the bisector t of the angle C, and the segments p and q of the side c made by t.

Proof. 1. Circumscribe a circle about the

bisector CD to meet the circumference in E.

ABC. Produce the
Draw EB, and let

426. From the above relation and from § 338 the formula for the length of the bisector to upon any side of a triangle, as c, is derived as follows:

Solving for te,
But from § 338,

and

t = ab - pq.

p: q = b : a.

..p :p + q = b : a + b,

Why?