If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the internal segments of the base shall have to one another the same ratio as the other sides of the triangle have. Conversely: If the internal segments of the base have to one another the same ratio as the other sides of the triangle have, the straight line drawn from the vertex to the point of section shall bisect the vertical angle. (1) Let the vertical ▲ BAC of the ▲ ABC be bisected by AD, which meets the base at D: it is required to prove that BD: DC = BA: AC. Through C draw CE || DA, and let CE meet BA produced at E. I. 31 Because DA is || CE, a side of the ▲ BCE, (2) Let BD: DC = BA: AC, and AD be joined: it is required to prove L BAD Through C draw CE || DA, = L DAC. and let CE meet BA produced at E. Because DA is || CE, a side of the ▲ BCE, and AE = AC, LAEC: = LACE. But because DA and CE are parallel, .*. LAEC = L BAD, and ACE: = L DAC; .. L BAD = L DAC. I. 31 VI. 2 Hyp. V. 11 V. 9 I. 5 I. 29 1. With the same figure and construction as in I. 10, prove that AB is bisected. 2. If a straight line bisect both the base and the vertical angle of a triangle, the triangle must be isosceles. 3. The bisector of an angle of a triangle divides the triangle into two others, which are proportional to the sides of the bisected angle. 4. ABC is a triangle whose base BC is bisected at D; Ls ADB, ADC are bisected by DE, DF meeting AB, AC at E, F. Prove EF BC. 5. Trisect a given straight line. 6. Divide a given straight line into parts which shall be to one another as 3 to 2. 7. Divide a given straight line into n equal parts. 8. The bisectors of the angles of a triangle are concurrent. 9. Express BD and DC (fig. to the proposition) in terms of a, b, c, the three sides of the triangle. 10. AB is a diameter of a circle, CD a chord at right angles to it, and E any point in CD; AE, BE produced cut the circle at Fand G. Prove that the quadrilateral CFDG has any two of its adjacent sides in the same ratio as the other two. 11. H is the middle point of BC (fig. to the proposition): prove HC: HD BA + AC : BA AC. 12. The straight lines which trisect an angle of a triangle do not trisect the opposite side. If the exterior vertical angle of a triangle be bisected by a straight line which also cuts the base produced, the external segments of the base shall have to one another the same ratio as the other sides of the triangle have. Conversely: If the external segments of the base have to one another the same ratio as the other sides of the triangle have, the straight line drawn from the vertex to the point of section shall bisect the exterior vertical angle. F E B (1) Let the exterior vertical ▲ CAF of the ▲ ABC be bisected by AD, which meets the base produced at D: it is required to prove that BD: DC = BA: AC. Through C draw CE || DA, and let CE meet BA at E. Because DA and CE are parallel, L FAD = AEC, and ▲ DAC: = LACE. But FAD = L DAC; I. 31 I. 29 I. 6 Because DA is || CE, a side of the ▲ BCE, * Assumed in Pappus, VII. 39, second proof. = (2) Let BD: DC – BA : AC, and AD be joined : VI. 2 V. 7 Because DA is || CE, a side of the ABCE, VI. 2 Hyp. But because DA and CE are parallel, L AEC = ▲ FAD, and ▲ ACEL DAC; V. 11 V. 9 I. 5 I. 29 1. What does the proposition become when the triangle is isosceles? 2. The bisector of the vertical angle of a triangle, and the bisectors of the exterior angles below the base, are concurrent. 3. Express BD and DC (fig. to the proposition) in terms of a, b, c, the three sides of the triangle. 4. Prove the tenth deduction from VI. 3 when E is taken in CD produced. 5. P is any point in the Oce of the circle of which AB is a diameter; PC, PD drawn on opposite sides of AP, and making equal angles with it, meet AB at C and D. Prove AC: CB AD: DB. = 6. AB is a straight line, and C is any point in it; find in AB produced a point D such that AD: DB = AC: CB. 7. Prove the proposition by cutting off from BA produced, AE = AD, and joining DE. 8. If in any ▲ ABC there be inscribed a ▲ XYZ (X being on BC, Y on CA, Z on AB), such that every two of its sides make equal angles with that side of ▲ ABC on which they meet, then AX, BY, CZ are respectively 1 BC, CA, AB. Examine the case when X and Y are on BC and AC produced. PROPOSITION 4. THEOREM. If two triangles be mutually equiangular, they shall be similar, those sides being homologous which are opposite to equal angles.* In As ABC, DCE, let ▲ ABC = LDCE, BCA it is required to prove As ABC, DCE similar. Place A DCE so that CE may be contiguous to BC, and in the same straight line with it. ▲ ABC + ↳ DEC is less than 2 rt. 4s; = I. 22 *This theorem is usually attributed to Thales (640-546 B.C.). |