Definitions. (V.1). A magnitude is a part of a magnitude, the less of the greater, when it measures
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1 hapter 8 Euclid s Elements ooks V 8.1 V.1-3 efinitions. (V.1). magnitude is a part of a magnitude, the less of the greater, when it measures the greater. (V.2). The greater is a multiple of the less when it is measured by the less. Euclid. (V. 1). If there be any number of magnitudes whatever which are, respectively, equimultiples of any magnitudes equal in multitude, then, whatever multiple one of the magnitudess is of one, that multiple also will all be of all. ma 1 + ma ma n = m(a 1 + a a n ). (V. 2). If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth. If a 1 = ka 2, a 3 = ka 4, and a 5 = ma 2, a 6 = ma 4, then a 1 + a 5 =(k + m)a 2, a 3 + a 6 =(k + m)a 4. (V. 3). If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will also be equimultiples respectively, the one of the second and the other of the fourth. If a 1 = ka 2, a 3 = ka 4, then ma 1 = Ka 2, ma 3 = Ka V.4-7 efinitions. (V.3). ratio is a sort of relation in respect of size between two magnitudes of the
2 302 Euclid s Elements ooks V same kind. (V.4). Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another. Given a, b, there exist integers m and n such that ma > b and nb > a. (V.5). Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order. a : b = c : d if for ma > kb, mc > kd, orma = kb, mc = kd, orma < kb, mc < kd. (V.6). Let magnitudes which have the same ratio be called proportional. (V.11). The term corresponding magnitudes is used of antecedents in relation to antecedents, and of consequents in relation to consequents. Euclid (V.4). If a first magnitude have to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. If a : b = c : d, then ma : kb = mc : kd. (V. 5). If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, then the remainder will also be the same multiple of the remainder that the whole is of the whole. ka kb = k(a b). (V. 6). If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal the same or equimultiples of them. If a = kc, b = kd, then a mc = b md or a mc = Kc, b md = Kd. (V. 7). Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes. If a = b, then a : c = b : c. If a = b, then d : a = d : b. 8.3 omparison of ratios efinition (V.7). When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth. a : b>c: d if ma > kb, mc kd.
3 8.4 Operations of ratios 303 Euclid (V.8). Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. If a>b, then a : c>b: c, and d : b>d: a. (V. 9). Magnitudes which have the same ratio to the same equal one another; and magnitudes to which the same has the same ratio be equal. If a : c = b : c, then a = b. Ifd : a = d : b, then a = b. (V. 10). Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. If a : c>b: c, then a>b.ifd : a>d: b, then a<b. Euclid. (V. 11). Ratios which are the same with the same ratio are also the same with one another. Transitive law: If a : b = c : d and c : d = e : f, then a : b = e : f. (V. 12). If any number of magnitudes be proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. If a 1 : b 1 = a 2 : b 2 := = a n : b n, then a k : b k =(a 1 + a a n ):(b 1 + b b n ). (V. 13). If a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth. a : b = c : d and c : d>e: f, then a : b>e: f. (V. 14). If a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. Let a : b = c : d. Ifa>, =, <c, then b>, =, <d. (V. 15). Parts have the same ratio as the same multiples of them taken in corresponding order. a : b = ma : mb. 8.4 Operations of ratios efinitions (V.12). lternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent.
4 304 Euclid s Elements ooks V If a : b = c : d, the proportion a : c = b : d is obtained alternatively. (V.13). Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent. The inverse ratio of a : b is b : a. (V.14). omposition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself. The composition of the ratio a : b is a + b : b. (V.15). Separation of a ratio means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself. The separation of the ratio a : b is a b : b. (V.16). onversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent. The conversion of the ratio a : b is a : a b. Euclid (V. 16). If four magnitudes be proportional, then they be also proportional alternately. If a : b = c : d, then a : c = b : d. (V. 17). If magnitudes be proportional componendo, they will also be proportional separando. If a : b = c : d, then a b : b = c d : d. (V. 18). If magnitudes be proportional separando, they will also be proportional componendo. If a : b = c : d, then a + b : b = c + d : d. (V. 19). If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole. If a : b = c : d, then a c : b d = a : b. Porism. If magnitudes be proportional componendo, they will also be proportional convertendo. If a : b = c : d, then a : a b = c : c d. Propositions V.15,17,18,19 can be combined as follows. If a : b = c : d, then for integers h, k, m, n, one has ha + kb : ma + nb = hc + kd : mc + nd.
5 8.5 Proportions Proportions efinitions (V.8). proportion in three terms is the least possible. (V.17). ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two be in the same proportion, as the first is to the last among the first magnitudes, so is the first to the last among the second magnitudes. Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms. Given a, b, c and d, e, f with a : b = d : e and b : c = e : f, the ratio a : c = d : f is obtained ex aequali. (V.18). perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, antecedent is to consequent among the first magnitudes as antecedent is to consequent among the second magnitudes, while, the consequent is to a third among the first magnitudes as a third is to the antecedent among the second magnitudes. Given a, b, c and d, e, f with a : b = e : f and b : c = d : e, the ratio a : c = d : f is obtained ex aequali in perturbed proportion. Euclid. (V. 20). If there be three magnitudes, and others equal to them in multitude, which taken two and two be in the same ratio, and if ex aequali the first is greater than the third, then the fourth is also greater than the sixth; if equal, equal, and; if less, less. Given a, b, c and d, e, f with a : b = d : e and b : c = e : f, if a>, =, <c, then d>, =, <f. (V.22). If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together be in the same ratio, they be also in the same ratio ex aequali. If a, b, c,...,g, h, and,,,...g, H be such that a : b = :, and b : c = :,...,g : h = G : H, then a : h = : H. Euclid (V.21). If there be three magnitudes, and others equal to them in multitude, which taken two and two together be in the same ratio, and the proportion of them is perturbed, then, if ex aequali the first magnitude is greater than the third, then the fourth is also greater than the sixth; if equal, equal; and if less, less. Given a, b, c and d, e, f with a : b = e : f and b : c = d : e, if a>, =, <c, then d>, =, <f. (V. 23). If there be three magnitudes, and others equal to them in multitude, which taken two and two together be in the same ratio, and the proportion of them be perturbed, then they be also in the same ratio ex aequali. Given a, b, c and d, e, f, ifa : b = e : f and b : c = d : e, then a : c = d : f.
6 306 Euclid s Elements ooks V Euclid (V.24). If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth. If a : b = c : d and e : b = f : d, then a + e : b = c + f : d. (V.25). If four magnitudes be proportional, then the sum of the greatest and the least is greater than the sum of the remaining two. If a : b = c : d, and a>b, a>c, then a is the greatest and d is the least, and a + d>b+ c.
7 hapter 9 Euclid s Elements ooks VI 9.1 VI.1-2 Euclid (VI.1). Triangles and parallelograms which are under the same height are to one another as their bases. E F H G Given: Triangle and. To prove: : = :. onstruction: Multiples of H = m and L = k of the bases. Proof : H = m and L = k. If m >, =, <k, H >, =, <L, H >, =, < L, m >, =, <k. Therefore, : = :. Euclid (VI.2). If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle. Given: Triangle with on and E on such that E//. To prove: : = E : E. K L
8 308 Euclid s Elements ooks VI E Proof : : = : (VI.1) : + = : +, (V.14) : = :. Similarly, from E : E = E : E, E : = E :. Since E is parallel to, triangle = E. (I.37) Therefore, : = E :, (V.7) : = E : (V.11) and : = E : E (V.17). Given: Triangle with on and E on such that : = E : E. To prove: E//. Proof : : = : (VI.1) : + = : +, (V.14) : = :. Similarly, from E : E = E : E, E : = E :. If : = E : E, then : = E : (V.) Therefore, : = E :, and = E (V.9) These two triangles have the same base and equal areas. E and are parallel. (I.39). 9.2 The angle bisector theorem Euclid (VI.3). If an angle of a triangle is bisected and the straight line cutting the angle cuts the base also, the segments of the base will have the same ratio of the remaining sides of the triangle; and if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of the section will bisect the angle of the triangle. Given: Triangle and the bisector of angle. To prove: : = :.
9 9.3 Similar triangles 309 E onstruction: The parallel of through to intersect produced at E. Proof : Since //E, E = (I.29) = ( bisects angle ) = E (I.29) Therefore, E =. (I.5) Now, : = : E. (VI.2) We have : = :. onverse: exercise. 9.3 Similar triangles efinition (VI.1). Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional. Euclid (Euclid VI.4). In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. If in triangles and XY Z, = X and = Y, 1 then : XY = : YZ = : ZX. (VI.5). If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend. onverse of VI.4. Euclid (Euclid VI.6). If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend. 1 It is enough to assume the equality of two pairs of angles. The equality of the third pair follows from I.32.
10 310 Euclid s Elements ooks VI If in triangles and XY Z, = X and : = XY = Y and = Z. : XZ, then (VI.7). If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional. If in triangles and XY Z, = X, : XY = : YZ, and, Z both acute or non-acute, then = Y and = Z. The triangles are equiangular and : XY = : YZ = : XZ. Euclid (VI.8). If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another. Porism. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Proof. Triangles and are equiangular. They are similar by VI.4. Similarly for the pairs, and,. It follows that : = :, and is a mean proportional of and. Euclid (VI.9). From a given straight line to cut off a prescribed part. To divide a given segment into m parts and n parts. (VI.10). To cut a given uncut straight line similarly to a given cut straight line. Given two lines and XY Z, to construct in such that : = XY : YZ. (VI.11). To two given straight lines to find a third proportional. Given a and b,tofindc such that a : b = b : c. (VI.12). To three given straight lines to find a fourth proportional. Given a, b, c, to find d such that a : b = c : d. (VI.13). To two given straight lines to find a mean proportional. Given a and b,tofindm such that a : m = m : b.
11 9.4 Similar figures Similar figures Euclid (VI.14). In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. H E F Given: F and EG are equal and equiangular parallelograms. To prove: : E = : F. onstruction: omplete the parallelogram EHF. Proof : F : FEH = : E; (VI.1) EG : FEH = : F, (VI.1) Since F = EG,wehave : E = : F. onverse: exercise. Euclid (VI.15). In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. (VI.16). If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportional. (VI.17). If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes is equal to the square on the mean, the three straight lines are proportional.
12 312 Euclid s Elements ooks VI Euclid (VI.18). On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilinear figure. efinitions. (V.9). When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second. (V.10). When four magnitudes are [continuously] proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion. Euclid (VI.19). Similar triangles are to one another in the duplicate ratio of the corresponding sides. (VI.20). Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. Porism. Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. nd it was also proved in the case of triangles; therefore, also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Euclid (VI.21). Figures which are similar to the same rectilineal figure are also similar to one another. Euclid (VI.22). If four straight lines are proportional, the rectilineal figures similar and similarly described upon them are also proportional; and if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional. Euclid (VI.23). Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides. Euclid (VI.24). In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another. Euclid (VI.25). To construct one and the same figure similar to a given rectilineal figure and equal to another rectilineal figure.
13 9.5 pplication of areas pplication of areas Euclid (VI.26). If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole. G K E F H Given: Similar parallelograms and EF G with E and G on and respectively, i.e., : E = : G. To prove: F lies on the diagonal. Proof (by contradiction): Suppose F does not lie on the diagonal. One of the lines EF and GF, say GF, intersects the diagonal at a point H. omplete the parallelogram GHK (with K on ). and EF G are similar and similarly situated parallelograms. : E = : G. Since GH//, //, and HK//, //, GH// and HK//. We have : K = : H = : G. (VI.2) Therefore, : K = : G. It follows that : E = : K, and E = K. Therefore GF = E = K = GH, and the points F and H coincide. This contradicts the hypothesis that F does not lie on the diagonal. Euclid (VI.27). Of all the parallelograms applied to the same straight line and deficient by parallelogramic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.
14 314 Euclid s Elements ooks VI Euclid (VI.28). To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogramic figure similar and similarly situated to a given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect. H G P F K N T O Q R E S Given a line a and a parallelogram with sides in the ratio b : c and area, to find x such that (a bx) cx =. L M Given: segment, a rectilineal area, and a parallelogram. To construct: parallelogram SQT with S between and, an angle equal to one of the parallelogram, such that (i) the area of SQT is equal to, (ii) the defect QSR is similar to. onstruction: Let E be the midpoint of. onstruct a parallelogram GEF similar to the given parallelogram. omplete the parallelogram F H. If the parallelogram HEG has the same area as, we are done. If HEG and GEF have areas greater than that of, construct a parallelogram KLMN with area equal to the excess of GEF over. Let GOQP be the parallelogram with O on GE, P on GF, and area equal to that of KLMN. Then Q is on the diagonal G (VI.26), and SQT = ESQO+EOT = ESQO+ERO = OSQO+PSF = GEF GOQP, which has the same area as. In the special case of a square, this is to construct, for a given length a and a given area, a length x such that (a x)x =. onstruction: Let M be the midpoint of. onstruct the perpendicular bisector of. onstruct a circle, center, radius a side of the square of area, to intersect the perpendicular bisector of at P. onstruct a square MPQR with Q and R on the same side of. R is the point such that the rectangle formed by R and R is equal to the given area, and is deficient by a square on R.
15 9.5 pplication of areas 315 Euclid (VI.29). To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogramic figure similar to a given one. F L M K H E P N Q O G Euclid (VI.31). In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Euclid (VI.32). If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line. E Given: Triangles and E with //, //E, and : = : E. To prove:,, E are collinear. Proof : Note that = = E. Since also : = : E, the triangles and E are equiangular. (VI.7) + + E = + + =2rt. s. This shows that,, E are collinear. (I.14) Euclid (VI.33). In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centers or at the circumferences.
16 316 Euclid s Elements ooks VI 9.6 The regular pentagon and division in extreme and mean ratio efinition (V.3). straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less. Given a line, to construct E on such that : E = E : E. Euclid (VI.30). To cut a given finite straight line in extreme and mean ratio. F H M E K Given: line. To construct: point E on such that : E = E : E, i.e., ( + E) E =. onstruction: a square H with the midpoint M of, a circle, center M, radius M, to intersect produced at K, a square KE with E in. Euclid (XIII.1). If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half. (XIII.2). If the square on a straight line is five times the square on a segment on it, then, when the double of the said segment be cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. (XIII.3). If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment. (XIII.4). If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together is triple of the square on the greater segment. (XIII.5). If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. Euclid (XIII.7). If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon is equiangular.
17 9.6 The regular pentagon and division in extreme and mean ratio 317 P E Given: Equilateral pentagon E with = =. To prove: = E = = =. onstruction: The diagonals and let, intersect at P. Proof : Triangles E,, and are congruent since they are isosceles triangles with equal vertical angles and equal adjacent sides. (SS) Therefore, the six (solid) marked acute angles are all equal. Triangles P and P are congruent. (S) Therefore, P = P, and triangles P E and PE are congruent. (SSS) PE = PE. From this, =. Triangle E is congruent to the three isosceles triangles, and the two marked angles (dotted) are equal to the sid marked angles. Finally, triangles E and E are congruent. (SSS) E = E. In the pentagon, E =. The pentagon is equiangular. Given: Equilateral pentagon E with = =. To prove: = E = = =. Proof : Exercise. Euclid (XIII.8). If in an equilateral and equilateral pentagon straight lines subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon. Euclid (XIII.9). If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. onsider a regular pentagon E inscribed in a circle, center O. Let F be the midpoint of the arc (so that F is a side of a regular decagon inscribed in
18 318 Euclid s Elements ooks VI F G E O the circle). Extend F and O to intersect at G. (1) Triangle OF is isosceles, with angles = F =2 O. (2) Triangle GO is also isosceles with O = =2 G. Therefore, F divides G in the mean and extreme ratio. (IV.10) Note also that GF = FO = s 6. Euclid (XIII.10). If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the sides of the hexagon and on that of the decagon inscribed in the same circle. Euclid (XIII.11). If in a circle which has its diameter rational an equilateral pentagon is inscribed, the side of the pentagon is the irrational straight line called minor. Proof. Let E be a regular pentagon inscribed in a circle (O). Let F be the midpoint of the arc so that F is a side of a regular decagon inscribed in the same circle. Let P be the midpoint of the arc F (so that P is a side of a regular 20-gon inscribed in the circle). Join OP to intersect at H. P F H E O (1) Since F is the midpoint of the arc, triangle F is isosceles. The radius OP is the perpendicular bisector of F. Since H lies on this radius, it is
19 9.6 The regular pentagon and division in extreme and mean ratio 319 equidistant from and F. The triangle HF is also isosceles. The two triangles F and HF, sharing a common base angle at, are similar. Therefore, : F = F : H, and H = F 2 = s (2) Triangle O is clearly isosceles, with angles 72, 54, and 54. In triangle HO, clearly =54. lso, OH =54, being 3 of O. Therefore, it 4 is also isosceles with the same base angle as triangle O. From the similarity of O and HO,wehave : O = O : H, and H = O 2 = s 2 6. (3) The result follows from noting that H + H = 2 = s 2 5. Euclid (XIII.12). If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle.
20 320 Euclid s Elements ooks VI 9.7 onstruction exercises for ooks VI 1. Inscribe a square in a given segment of a circle. 2. Inscribe a square in the intersection of two (equal) circles whose centers lie on each other. 3. onstruct a triangle given the lengths of two sides and the bisector of the angle between the two given sides terminated by the third. 4. Given triangle with, construct an equal isosceles triangle XY with X on and Y on. 5. Inscribe a square in a regular pentagon.
21 hapter 10 Modern reorganization of Euclid s ook VI Theorem 70 (Euclid VI.1) If two triangle have equal altitudes, then the ratio of their areas is equal to the ratio of their bases. Theorem 71 (Euclid VI.2) straight line parallel to a side of a triangle divides the remaining two sides, extended if necessary, in the same ratio. Theorem 72 (Euclid VI.2) If two sides of a triangle are divided in the same ratio, both internally or both externally, the line joining the points of division is parallel to the third side. Theorem 73 (Euclid VI.3) (ngle bisector theorem) The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If X and X respectively the internal and external bisectors of angle, then X : X = c : b and X : X = c : b. Z c Z b X X
22 322 Modern reorganization of Euclid s ook VI Theorem 74 (Euclid VI.3) (ngle bisector theorem) (a) If X is a point on the side such that X : X = :, then the line X is the (internal) bisector of angle. (b) If X is a point on the extension of such that X : X = :, then the line X bisects angle externally. asic onstruction 19 (Euclid VI.9) To divide a given line segment in a given ratio, internally and externally. asic onstruction 20 (Euclid VI.12) To construct the fourth proportional of three given line segments. Theorem 75 (Euclid VI.4) If two triangles are equiangular, their corresponding sides are proportional. Theorem 76 (Euclid VI.5) If the three sides of one triangle are proportional to the three sides of a second triangle, then the two triangles are equiangular. Theorem 77 (Euclid VI.6) If two triangle have one pair of equal angles, and if the sides containing these angles are proportional, then the two triangles are equiangular. Theorem 78 (Euclid VI.8) If a perpendicular is drawn from the right angle of a right-angled triangle to the hypotenuse, the triangles on each side of the perpendicular is similar to the whole triangle, and to one another.
23 323 Theorem 79 (Euclid VI.19) The ratio of the areas of two similar triangles is equal to the ratio of the squares of corresponding sides. Theorem 80 If a segment is divided externally at X, and if is a point (not on the line ) such that X X = X 2, then the circle is tangent to X at. O X asic onstruction 21 (Euclid VI.13) To construct the mean proportional of two given line segments. asic onstruction 22 (Euclid VI.30) To divide a segment in the golden ratio, e.g., to construct a point P such that P = P 2. asic onstruction 23 (Euclid IV.10) To construct a triangle with = =2.
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