METHODS OF PROOF 245. Superposition. Two geometric figures may be shown to be equal or unequal by superposition; that is, by properly placing one upon the other. Thus, AABC is equal to ADEF, if they coincide throughout, when one is placed upon the other. B E F In theoretical superposition, the process is imagined to be done, as in the following theorems. 246. Congruence. If two geometric figures can be shown to coincide by superposition, they are equal in the sense that they are equal in every respect, or congruent. Figures which are congruent have the same size and the same shape. 247. Theorem. Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. Given ABC and DEF, having AB-DE, AC=DF, and ZA = ZD. To prove AABC = ADEF. Proof. Place AABC upon ADEF so that A falls on ZD and AC falls along DF. The point C will fall on the point F. (Since it is given that AC=DF.) AB will fall along DE. (Since it is given that ZAZD.) 248. Theorem. Two triangles are congruent if two angles and the included side of one are equal respectively to two angles and the included side of the other. Given ABC and DEF, having ZA ZD, ZC=ZF, and AC=DF. To prove AABC ADEF: Proof. Place AABC upon ADEF so that AC falls on DF. Also BC will fall along EF, since ≤C = ZF. Since two straight lines can intersect in only one point, B must fall on E. .. AABCDEF. § 246 249. Theorem. Two parallelograms are congruent if they have two adjacent sides and the included angle of one equal respectively to two adjacent sides and the included angle of the other. Given parallelograms ABCD and EFHK, having AB=EF, AD=EK, and ZA = ZE. ABCD upon □EFHK so that AD coincides 250. Indirect Proof. When the indirect proof of a proposition is used, the line of reasoning begins by supposing that the conclusion of the proposition is wrong. Then it is shown that this supposition leads to a statement which is not true. This proves that the supposition is false and therefore, that the original proposition is true. This method of proof is known as the reductio ad absurdum (reduction to an absurdity). 251. Theorem. If two angles of a triangle are unequal, the sides opposite these angles are unequal, the greater side being opposite the greater angle. Given <A><C in AABC. To prove BC>BA. Proof. It must be true that BC BA or BC <BA or BC>BA. Suppose BC=BA. Then ZA = 2C. Suppose BCBA. Then ZA <ZC. A B Why? Why? But, since it was given that ZA><C, neither of these suppositions can be true. ... BC>BA. 252. Theorem. Only one perpendicular can be drawn to a given line at a given point in the line. Given DC LAB at C. To prove that no other perpendicular can be drawn to AB at C. Proof. Suppose another line EC LAB at C. Then ZECB and ZDCB are both right angles. DE B But, since all right angles are equal, ZECB and ZDCB cannot both be right angles. Therefore the supposition that EC LAB is not true and DC is the only perpendicular which can be drawn to AB at C. 253. Theorem. Only one perpendicular can be drawn to a given line from a given external point. Given EC LAD from E. To prove that no other perpendicular can be drawn from E to AD. Proof. Suppose another line EBLAD from E. C E Why? But this is impossible since a triangle can have only one right angle. Then the supposition that EB 1 AD is not true. .. EC is the only perpendicular that can be drawn to AD from E. 254. Theorem. If two parallel lines are cut by a transversal, the alternate interior angles are equal. H E N B M Given CD and EF, two parallel lines cut by the transversal AB. To prove CMN= <MNF. Proof. Suppose ZCMN is not equal to ZMNF. Also suppose a line HK to be drawn through N, making ZMNK=ZCMN. Then, through N, there are two lines parallel to CD, which is impossible. Why? Hence the supposition that ZCMN is not equal to ZMNF is not true. :. ¿CMN= <MNF. 255. Theorem. If two parallel lines are cut by a transversal, the exterior interior angles are equal. Prove by the indirect method, using the figure of § 254. 256. Theorem. If two parallel lines are cut by a transversal, the consecutive interior angles are supplementary. Prove by the indirect method. 257. Theorem. If two parallel lines are cut by a transversal, the alternate exterior angles are equal. Prove by the indirect method. 258. Theorem. If two triangles have two sides of one equal respectively to two sides of the other but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second Given triangles DEF and ABC having DE=AB, EF=BC, and <E><B. To prove that DF>AC. Proof. Place ABC upon ADEF so that AB coincides with DE and AABC takes the position of ADEK. But Then Or Bisect ZKEF by the line EH and draw KH. Then ΔΗΕΚ = ΔΗΕΡ DH+HK>DK DH+HF>DK HK=HF Why? Why? Why? Why? |
