Proof. Angle F G H is equal to angle E (by construction). Angle KHM is equal to angle E (by construction). .. by Axiom 1, angle F G H is equal to angle K H M. To each of these equals add the angle KH G. angles FGH, G H K are equal to angles GHK, KH M. But, by Euc. I. 29, angles FGH, GHK are equal to two right angles. ... angles GHK, KH M are equal to two right angles. Hence, by Euc. I. 14, G H is in the same straight line with H M. Again, FK is parallel to GM, and angle FKH is equal to interior opposite angle KHM. To each of these add the angle HKL. Then, by Axiom 2a, angles M HK, HKL are equal to angles HKL, HK F. But, by Euc. I. 29, MHK, HKL are equal to two right angles. ... HKL, HKF are equal to two right angles. Hence, by Euc. I. 14, F K is in the same straight line with K L. Again, FG is parallel to KH, and LM is parallel to KH. .. by Euc. I. 30, F G is parallel to L M, and FL is parallel to GM. FGML is a parallelogram. Again, the triangle A D B is equal to the parallelogram FH (by construction), and the triangle DB Cis equal to the parallelogram KM (by construction). .. by Axiom 2, The rectilineal figure ABCD is equal to the parallelogram FGML, which has been described with its angle FG M equal to the angle Repeat. The enunciations of Euc. I. 29 and 34, Axiom 1, and definition of a square. General Enunciation. To describe a square on a given straight line. Particular Enunciation. Given. The straight line A B. A Required. To describe a square on A B. B Construction. (a) From the point A draw A Cat right angles to AB (by Euc. I. II). (b) Make A D equal to AB (by Euc. I. 3). (c) Through D draw DE parallel to A B (by Euc. I. 31). (d) Through B draw BE parallel to AB (by Euc. I. 31). ADEB shall be a square. Proof. Г By (c) and (d), A DEB is a parallelogram. .. by Euc. I. 34, (e) A B is equal to D E, and AD to BE. But by (b), AB is equal to AD. by Axiom 1, (f) A B, BE, E D, and D A are all equal. Again, A B is parallel to D E, and AD meets them. B (g).. angles BAD, ADE are equal to two right angles. But BAD is a right angle. .. ADE is also a right angle. And by Euc. I. 34, ABE, BED are right angles. (h). BAD, ADE, ABE, BED are all right angles. Hence, by (f) and (h), The figure ABED is a square, and it is described on the line A B. Q. E. F. Given. ABC a right- (b) EB, BC are in the same (c) AF is equal to E C. (Use Euc. I. 15, then I. 14, to prove (a) and (b), and the definition of a square to prove (c).) THEOREM (Euclid I. 47). Repeat. The enunciations of Euc. I. 4, 14, 41; Axioms 2, 6a, and 11, and definition of a square. General Enunciation. In any right-angled triangle, the square which is described on the side subtending the right angle, is equal to the squares on the sides which contain the right angle. Particular Enunciation. Given. The right-angled triangle ABC: of which (a) the angle BAC is the right angle. Required. To prove that the square described on Fig. 1 B BC shall be equal to the squares described on BA, A C. Construction. On BC (by Euc. I. 46) describe the square BCDE. (b) Through the point A, draw AL (Fig. 3) parallel to BD and CE (by Euc. I. 31). Join AD, FC. Proof. BAC is a right angle (a). BAG is a right angle (definition). by Euc. I. 14, (c) CA is in the same straight line with AG. BAC is a right angle (a). CAH is a right angle (definition). |