THEOREM XI. The rectangle contained by the sum of two lines and their difference is equal to the difference A C (a - b) × a Produce AB to A', making BA' = BA. b2 (a-t) xb В Then AC-AB-BC, and A'C= AB+ BC, therefore the rectangle AC. A'C=square on AB - square on BC. Again, produce AB to C' making BC' =BC. = Then AC AB - BC, and AC' = AB+BC, therefore the rectangle AC. AC' the square on AB-square on BC. = A C B C These forms of the theorem are often employed. THEOREM XII. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. Let ABC be a right-angled triangle, having the right angle BAC: the square described on the side BC shall be equal to the squares described on the sides BA, AC. On BC describe the square BDEC, and on BA, AC describe the squares GB, HC: through A draw AL parallel to BD or CE: and join AD, FC. Then, because the angle BAC is a right angle, and that the angle BAG is also a right angle, the two straight lines AC, AG on the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles: Therefore CA is in the same straight line with AG. For the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle FBA, for each of them is a right angle. Add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC. And because the two sides AB, BD are equal to the two sides FB, BC each to each; and the angle DBA is equal to the angle FBC, therefore the triangle ABD is equal to the triangle FBC. Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD and between the same parallels BD, AL. And the square GB is double of the triangle FBC, because they are on the same base FB and between the same parallels FB, GC. GB. But the doubles of equals are equal to one another. Therefore the parallelogram BL is equal to the square In the same manner by joining AE, BK, it can be shewn that the parallelogram CL is equal to the square CH. Therefore the whole square BDEC is equal to the two squares GB, HC. And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square described on the side BC is equal to the squares described on the sides BA, AC. Wherefore, in any right-angled triangle, &c. Q. E. D. The proposition may also be demonstrated by showing that the square on the hypothenuse may be divided into parts capable of being so placed as to coincide with the squares on the sides. The figure shews the necessary construction. We shall give further on a full discussion of this important proposition. THEOREMS. 1. THE diagonals divide a parallelogram into four equal triangles. 2. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles contained by these sides supplementary, the triangles are equal. 3. If the middle points of the sides of a triangle be joined, the area of the triangle thus formed is one quarter of the area of the original triangle. 4. Every line which passes through the intersection of the diagonals of a parallelogram, divides the figure into two equal parts. 5. If a point within a parallelogram be joined to the vertices, the two triangles formed by the joining lines and two opposite sides are together equal to half the parallelogram. What is the case when the point is outside? 6. ABCD is a parallelogram; through E, F, &c. on AB, EH, FK, &c. are drawn parallel to BC: shew that the triangles AHE, EKF, &c. are together equal to half the parallelogram. 7. ABCD is a parallelogram, P a point outside it: the triangles PAB and PAD are together equal to the triangle PAC. What is the case if the point be inside? 8. The measure of a triangle is equal to half the product of the perimeter and the radius of the inscribed circle. 9. The area of a trapezium is equal to twice the area of the triangle formed by joining the extremities of one nonparallel side to the middle point of the other. 10. Being given the base and height of a trapezium of known area, to find the areas of the two triangles of which it is the difference. 11. The sum of the perpendiculars let fall on the sides of an equilateral triangle from any point within it is constant. |