3. Afterwards join DB; and name the figure which is double of the triangle DBC. 4. Show that DBC and EBC are halves of equal figures, and draw an inference by Ax. 7. 5. Join EG in the same figure, and show that EFG and EBC are halves of equal things ; also draw the inference. Triangles upon the same base and between the same parallels are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC. Then ABC shall be equal to DBC. Produce AD both ways to the points E, through B draw BE parallel to CA, (I. 31) and through o draw CF parallel to BD. F; W Then each of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal to DBCF, (I. 35) because they are upon the same base BC, and between the same parallels BC, EF. And because the diameter AB bisects the parallelogram EBCA, therefore the triangle ABC is half of EBCA. (I. 34) Also, since the diameter DC bisects the parallelogram DBCF, therefore the triangle DBC is half of DBCF; but the halves of equal things are equal ; (Ax. 7) therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q. E. D. 86. PROPOSITION XXXVIII.-THEOREM. Triangles upon equal bases and between the same parallels are equal to one another. Let the triangles ABC, DEF be upon the equal bases BC, EF, and between the same parallels BF, AD. Then ABC shall be equal to DEF. Produce AD both ways to G and H; through B draw BG parallel to CA, (I. 31) and through F draw FH parallel to ED. Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to one another, (I. 36) being upon equal bases and between the same parallels. And because AB bisects the parallelogram GBCA, therefore the triangle ABC is half of GBCA; (I. 34) and because DF bisects the parallelogram DEFH, therefore the triangle DEF is half of DEFH; but halves of equal things are equal; (Ax. 7) therefore the triangle ABC is equal to the triangle DEF. Wherefore, triangles upon equal bases, &c. Q. E. D. EXERCISE XLVII. 1. In the figure to XXXVII. prove the triangles ABE, DCF equal. 2. Also the triangles BAD, CAD. 3. In the figure of XXXVIII. let D coincide with A. Will the two triangles be between the same parallels ? 4. If the vertex of a triangle be joined to the middle of the base, show that the triangle is bisected. 5. If the base of a triangle be trisected, how may you trisect the triangle ? 6. Given that the diagonals of a parallelogram bisect one another, show, by XXXVIII., that they divide the figure into four equal triangles. 87. PROPOSITION XXXIX.-THEOREM. Equal triangles upon the same base and upon the same side of it are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it. B AD shall be parallel to BC. For, if it is not, through A draw AE parallel to BC, (I. 31) meeting BD, or BD produced, in E, and join EC. Then the triangle ABC is equal to the triangle EBC, (I. 37) because they are upon the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal to the triangle DBC; (Hyp.) therefore the triangle DBC is equal to the triangle EBC, the greater equal to the less, which is impossible; •therefore AE is not parallel to BC. In the same manner, it can be demonstrated, that no other line but AD is parallel to BC. AD is therefore parallel to BC. Wherefore, equal triangles, &c. Q. E. D. 88. PROPOSITION XL.—THEOREM. Equal triangles, upon equal bases in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the straight line BF, and on the same side of BF. Then they shall be between the same parallels. AD shall be parallel to BF. For, if it is not, through A draw AG parallel to BF, (I. 31) meeting ED, or ED produced, in G, and join GF. Then the triangle ABC is equal to the triangle GEF, (I. 38) because they are upon equal bases BC, EF, and between the same parallels BF, AG. But the triangle ABC is equal to the triangle DEF; (Hyp.) (.) therefore the triangle DEF is equal to the triangle GEF, the greater equal to the less, which is impossible. Therefore AG is not parallel to BF. And in the same way it may be demonstrated, that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D. 89. FURTHER Tests OF PARALLELISM. In addition to those we have observed in previous propositions, XXXIX. and XL. furnish most useful conditions of parallelism. It will be partly the aim of the next exercise to impress this view of them upon the memory. EXERCISE XLVIII. 1. Prove XL. with a figure in which AG meets ED produced. 2. What proposition may be thus enunciated ? If two triangles which have a base common, and are both on the same side of it, be equal, their vertices will lie upon a line which is parallel to the base. 3. Give an enunciation to XL. similar to this one. 4. If there be three or more equal triangles upon one base and on one side of it, prove that their vertices will be in one straight line. 5. In the triangle ABC, let D, E be middle points A of AB, AC. Show that the triangle BCD is half of ABC. 6. Similarly, show BEC is half of the same. 7. By what axiom are BDC, BEC equal ? Draw an inference with regard to the lines DE, BC. 8. Take F the middle point of BC, and show that, if EF be joined, it is parallel to AB, and that, F similarly, DF would be parallel to AC. B 90. PROPOSITION XLI.—THEOREM. If a parallelogram and a triangle be upon the same base and between the same parallels, the parallelogram is double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC, AE. Then the parallelogram ABCD shall be double of the triangle EBC Join AO. E B (I. 37) Then the triangle ABC is equal to EBO, because they are upon the same base BC, and between the same parallels BC, AE. |