In a right-angled triangle, the square described upon the side opposite to the right angle is equal to the squares described upon the sides which contain the right angle. Let ABC be a right-angled triangle, having the right angle BAC; the square described on BC is equal to the squares described upon the sides BA and AC. On the sides BC, BA, and AC describe the squares BE, BH, and CG (Prop. 46.); through A draw AN parallel to BD or CE, and join AE and BF. Then, because the angles BAC and GAC are two right angles, the lines E G le H B BA and AG form one right line (Prop. 14.); and because the angles ACF and BCE are equal (Ax. 11.), if вCA be added to each, the angles BCF and ACE are equal; and the sides BC and CF are equal to the sides EC and cA (Def. 31.); therefore the triangles BCF and ECA are equal (Prop. 4.): but the square CG is double of the triangle BCF, being on the same base CF and between the same parallels (Prop. 41.), and the parallelogram CN is double of the triangle ECA, being on the same base CE and between the same parallels (Prop. 41); therefore the square CG is equal to the parallelogram CN, being doubles of the equal triangles (Ax. 6.). In the same manner it can be shown that the square BH is equal to the parallelogram B N, by joining AD and CL; therefore the whole square BDEC is equal to the two squares CG and BH. Which was to be demonstrated. SCHOL. 1. In a right-angled triangle, the side opposite to the right angle is called the hypotenuse, and the other sides are called the base and perpendicular. SCHOL. 2. On this proposition is founded the rule in Mensuration for finding any side of a right-angled triangle when the other two sides are given. To find the hypotenuse:- - Add the square of the base to the square of the perpendicular,_and extract the square root of the sum. To find the base: - Take the square of the perpendicular from the square of the hypotenuse, and extract the square root of the difference. To find the perpendicular:- Take the square of the base from the square of the hypotenuse, and extract the square root of the difference. 1 t I S S PROP. XLVIII. THEOR. If the square described upon one side of a triangle be equal to the squares described upon the other two sides, the angle opposite to thut side is a right angle. In the triangle ABC, let the square of a C be equal to the squares of AB and BC; then the angle ABC is a right angle. From в draw BD perpendicular to BC (Prop. 11.), and equal to AB (Prop. 3.), and join CD. Because the angle DBC is a right angle, A the square of CD is equal to the squares of C B D BC and BD (Prop. 47.); but BD is equal to AB, therefore the square of CD is equal to the squares of BC and AB: but the square of a C is also equal to the squares of BC and AB (Const.); therefore the squares of AC and CD are equal (Ax. 1.), and therefore the side AC is equal to CD: and because AB is also equal to BD, and BC common, the triangles BAC and BDC are equal (Prop. 8.); therefore the angle ABC is equal to the angle DBC: but DBC is a right angle (Const.), therefore ABC is also a right angle. Which was to be demonstrated. EXERCISES. 1. Draw a perpendicular from the end of a line, without producing the line. 2. Show that every equilateral triangle is also equiangular. 3. If a line bisect the angle opposite to the base of an isosceles triangle, it bisects the base also. 4. Two perpendiculars cannot be drawn from the same point to the same right line. 5. The angles made by any number of lines meeting in a point are equal to four right angles. 6. All the interior angles of any rectilineal figure, are equal to twice as many right angles as the figure has sides, less four right angles. 7. The exterior angles of a rectilineal figure are equal to four right angles. 8. The parallelograms about the diagonal of a square are squares. 9. To bisect a right angle. 10. To bisect a right line. 11. The diagonals of a parallelogram divide it into four equal parts. 12. In a square the diagonals cross one another at right angles. 13. To bisect a triangle. 14. To bisect a square. 15. Given the base and perpendicular of a right-angled triangle, to find the hypotenuse. 16. To find a square equal to the sum of three squares. 17. To describe a square, when the diagonal is given. 18. If a triangle and a parallelogram be between the same parallels, and the base of the triangle be double the base of the parallelogram, the triangle and parallelogram are equal. 19. If two right lines be equal, their squares are equal. 20. Every parallelogram having one angle right, has all its angles right angles. THE END. LONDON: SPOTTISWOODES and SHAW, New-street-Square. |