155. THEOREM. If a series of parallel lines intercept equal segments on one transversal, they intercept equal segments on every transversal.

Given the parallel lines 4, 4, 3, 4, cutting the transversal t1 so that AB = BC= CD.

To prove that on the transversal t2 EF = FG = GH.

Proof: The figure ACGE is a parallelogram or a trapezoid according as t1t or not.

In either case BF, which bisects AC, also bisects EG (§ 152).

... EF FG.

=

Similarly, in the figure DHFB, FG = GH.

Outline of Proof: Through each vertex of the given triangle ABC draw a line parallel to the opposite side, forming a triangle A'B'C'.

Show that ACA'B, and AB'CB are parallelograms, and hence that B'C= AB = CA'. That is, C is the middle point of A'B'.

In the same manner show that A and B are the middle points of B'C' and 'c' respectively. Also show that the altitudes of ▲ ABC are the perpendicular bisectors of the sides of ▲ A'B'C', and therefore meet in a point (§132).

157. Definition. A segment connecting a vertex of a triangle with the middle point of the opposite side is called a median of the triangle.

158. THEOREM. The three medians of a triangle meet in a point which is two thirds the distance from each vertex to the middle point of its opposite side.

Given ▲ ABC with medians BD and AE meeting in O.

To prove that the median from C also passes through 0, and that 40 = AE, BO = BD, and CO = 3 CH.

Outline of Proof: Taking F and G, the middle points of OB and OA respectively, use §§ 151 and 144 to show that the figure GFED is a parallelogram, and hence that

DO = OF = FB and EO = OG = GA.

That is, o trisects AE and BD.

In the same way we find that AE and CH meet in a point o' which trisects each of them.

Hence O and o' are the same point. Therefore the three medians meet in a point which trisects each of them.

1. If one angle of a parallelogram is 120°, how many degrees in each of the other angles?

2. If the angles adjacent to one base of a trapezoid are equal, then those adjacent to the other base are equal.

3. If the angles adjacent to either base of a trapezoid are equal, then the non-parallel sides are equal and the trapezoid is isosceles. 4. In an isosceles trapezoid the angles adjacent to either base are equal.

5. Divide a segment AB into three equal parts. SUGGESTION. From A draw a segment AC, and on it lay off three equal segments, AD, DE, EF (§ 33).

D

FC

E

H G B

Draw FB and construct EG and DH each parallel to FB. Prove AH = HG = GB.

6. To cut braces for a roof, as shown in the figure, a carpenter needs to know the angle DBC when the angle DAB is given, it being given that AB = AD. Show how to find this angle. (See § 123, Ex. 14.)

7. If each of the perpendicular bisectors of the

sides of a triangle passes through the opposite vertex, what kind of a triangle is it? If it is given that two of the perpendicular bisectors of sides pass through the opposite vertices, what kind of a triangle is it? If only one?

8. Find the locus of the middle points of the segments joining a vertex of a triangle to all points on the opposite side.

9. Given a line 7 and a point P not in the line. Find the locus of the middle points of all segments drawn from P to l.

10. The length of the sides of a triangle are 12, 14, 16. Four new triangles are formed by connecting middle points of the sides of this triangle. What is the sum of the sides of these four triangles?

11. Draw any segment PA meeting a line 7 in A. Lay off AB on 1. With P and B as centers and with AB and AP as radii respectively, strike arcs meeting in C. Draw PC, and prove it parallel to l.

P

A

B

12. If the side of a triangle is bisected by the perpendicular upon it from the opposite vertex, the triangle is isosceles.

13. State and prove the converse of this theorem.

14. If in a right triangle the hypotenuse is twice as long as one side, then one acute angle is 60° and the other 30°.

SUGGESTION. Let D be the middle point of AB.

Use Ex. 12 and the hypothesis to show that ▲ ACD is A E

equilateral.

Prove the converse by drawing CD so as to make ≤BCD =

10 in.

LB.

B

15. A stairway leading from a floor to one 12 feet above it is constructed with steps 8 inches high and 10 inches wide. What is the length of the carpet required to cover the stairway, allowing 10 inches for the last step, which is on a level with the upper floor.

16. A stairway inclined 45° to the horizontal leads to a floor 15 feet above the first. What is the length of the carpet required to cover it if each step is 10 inches high? If each is 12 inches? If each is 9 inches?

2 feet

Can this problem be solved without knowing the height of the steps? Is it necessary to know that the steps are of the same height?

17. If the vertex angle of an isosceles triangle is 60°, show that it is equilateral.

18. By successively constructing angles of 60° divide the perigon about a point O into six equal angles. (This is possible because 360 ÷ 6 = 60.) With O as a center construct a circle cutting the sides of these angles in points A, B, C, D, F, F. Draw the segments AB, BC, etc. Show by Ex. 17 that each of the six triangles thus formed

is equilateral. Show also that ABCDEF is equiangular and equilateral, that is, AB = BC, etc., and ▲ ABC :

=

4 BCD, etc.

160. Definitions.

POLYGONS.

A polygon is a figure formed by a series of segments, AB, BC, CD, etc., leading back to the starting point A.

The segments are the sides of the polygon and the points A, B, C, D, etc., are its vertices. The angles A, B, C, D, etc., are the angles of the polygon.

A polygon is convex if no side when produced enters it. Otherwise it is concave.

Only convex polygons are here considered.

A polygon is equiangular if all its angles are equal and equilateral if all its sides are equal.

A polygon is regular if it is both equiangular and equilateral.

A segment connecting two non-adjacent vertices is a diagonal of the polygon.

The perimeter of a polygon is the sum of its sides.

161. THEOREM. The sum of the angles of a polygon having n sides is (2n−4) right angles.

Proof Connect one vertex with each of the other non-adjacent vertices, thus forming a set of triangles. Evidently the sum of the angles of these triangles equals the sum of the angles of the polygon.