PROPOSITION XXXVII. THEOREM. Triangles upon the same base, and between the same parallels, are equal to one another. Let As ABC, DBC be on the same base BC and between the same s AD, BC. Then must ▲ ABC= ▲ DBC. From B draw BE || to CA to meet DA produced in E. Then EBCA and FCBD are parallelograms, and EBCA=□ FCBD, I. 35. Now they are on the same base and between the same ||s. I. 34. and ▲ DBC is half of ☐ FCBD; I. 34. .. ▲ ABC= ▲ DBC. Ax. 7. Q. E. D. Ex. 1. If P be a point in a side AB of a parallelogram ABCD, and PC, PD be joined, the triangles PAD, PBC are together equal to the triangle PDC. Ex. 2. Two straight lines AB, CD intersect in E, and the triangle AEC is equal to the triangle BED. Shew that BC is parallel to AD. Ex. 3. If A, B be points in one, and C, D points in another of two parallel straight lines, and the lines AD, BC intersect in E, then the triangles AEC, BED are equal. PROPOSITION XXXVIII. THEOREM. Triangles upon equal bases, and between the same parallels, are equal to one another. E Let As ABC, DEF be on equal bases, BC, EF, and between the same s BF, AD. Then must A ABC= ▲ DEF. From B draw BG || to CA to meet DA produced in G. From F draw FH || to ED to meet AD produced in H. Then CG and EH are parallelograms, and they are equal, they are on equal bases BC, EF, and between the same Is BF, GH. Now ▲ ABC is half of □ CG, I. 36. and A DEF is half of EH; : AABC= ADEF. Ax. 7. Q. E. D. Ex. 1. Shew that a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts. Ex. 2. If the triangles in the Proposition are not towards the same parts, shew that the straight line, joining the vertices of the triangles, is bisected by the line containing the bases. Ex. 3. In the equal sides AB, AC of an isosceles triangle ABC points D, E are taken such that BD=AE. Shew that the triangles CBD, ABE are equal. PROPOSITION XXXIX. THEOREM. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal As ABC, DBC be on the same base BC, and on the same side of it. Join AD. Then must AD be || to BC. For if not, through A draw AO || to BC, so as to meet BD, or BD produced, in O, and join OC. As ABC, OBC are on the same base and between Then the same ||s, the less the greater, which is impossible; .. AO is not || to BC. I. 37. Нур. In the same way it may be shewn that no other line passing through A but AD is || to BC; .. AD is to BC. Q. E. D. Ex. 1. AD is parallel to BC; AC, BD meet in E; BC is produced to P so that the triangle PEB is equal to the triangle ABC: shew that PD is parallel to AC. Ex. 2. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equal, the quadrilateral has two opposite sides parallel. PROPOSITION XL. THEOREM. Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. E Let the equal As ABC, DEF be on equal bases BC, in the same st. line BF and towards the same parts. Join AD. Then must AD be || to BF. EF For if not, through A draw AO || to BF, so as to meet ED, or ED produced, in O, and join OF. Then ▲ ABC= ▲ OEF, :: they are on equal bases and between the same Is. I. 38. Нур. .. ▲ OEF=▲ DEF, the less the greater, which is impossible. .. AO is not || to BF. In the same way it may be shewn that no other line passing through A but AD is || to BF, .. AD is to BF. Q. E. D. Ex. 1. If the triangles be not towards the same parts, shew that the straight line joining the vertices of the triangles is bisected by the line containing the bases. Ex. 2. The straight line, joining the points of bisection of two sides of a triangle, is parallel to the base. Ex. 3. The straight lines, joining the middle points of the sides of a triangle, divide it into four equal triangles. H 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 ABCD and the A EBC be on the same base BC and between the same ||s AE, BC. Then must ABCD be double of ▲ EBC. Join AC. Then ▲ ABC= ▲ EBC, they are on the same base and between the same s ; I. 37. and ABCD is double of ▲ ABC, AC is a diagonal of ABCD ; .. ABCD is double of ▲ EBC. I. 34. Q. E. D. 1 Ex. 1. If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, the difference of the triangles thus formed is equal to half the parallelogram. Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogram. |