If the altitude be taken equal to twice the unit of length the linear measure of the base will be equal to the superficial measure of the triangle.

6. Find the area of a triangle the sides of which are 3, 31, and 11 inches.

4

7. Find the area of a triangle having sides equal to 5 and 4 inches containing an angle of 150 degrees.

8. To construct a triangle equal to a given polygon.
Let ABCDE be a polygon of five sides.

Join AC. Through B draw BF parallel to CA meeting EA produced in F. Join FC.

Then the triangle AFC= triangle ABC.

And therefore the quadrilateral figure FCDE= the pentagon ABCDE.

The given construction enables us to transform any polygon into an equal polygon with one side less. Thus every polygon may, by repetition of the process, be reduced to an equivalent triangle.

Hence if required its area may be found by construc

tion alone.

9. To construct a square equal to the sum of two given squares.

10. To construct a square equal to the difference of two given squares.

11. To divide a given line so that the sum of the squares of the parts may be equal to a given square. What are the necessary conditions?

12. To divide a given line so that the rectangle contained by the segments may be equal to a given square. The conditions?

13. Bisect a given triangle by a straight line drawn through a point in one of its sides.

BOOK IV.

PROPORTIONAL LINES. SIMILAR FIGURES.

THEOREM I.

PARALLEL straight lines which cut off equal parts from

'one of two straight lines cut off also equal parts from the other.

Let AB, AC be straight lines meeting at A, AD, DE equal parts cut off from AB; let DF and EG be parallel straight lines passing through D and E and meeting AC in Fand G.

Then shall AF = FG.

Draw DK parallel to FG.

D

A A

E

A

G

L

M

N 0

P

R

COR. 1. Similarly, if EL, LP, &c. be all equal to AD;

GM, MR, &c. shall be all equal to AF

COR. 2. From the congruent triangles ADF, DEK,

COR. 3. Therefore as many parts as there are in AL each equal to AD, so many are there in AM each equal to AF, and so many in LM each equal to DF.

THEOREM II.

If a straight line LM be drawn parallel to BC one of the sides of the triangle ABC,

Again, if AB contain AD n times, so that AB=n.AD,

then

AC-n. AF, and BC=n. DF.