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LESSONS IN GEOMETRY.-No. XXX.

LECTURES ON EUCLID,

(Continued from page 44.)
BOOK I.

PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles. Let A B C be a triangle, fig. 32, No. 1, and let one of its sides BC be produced to n. Then the exterior angle A CD is equal to the two interior and opposite angles CA B and ABC. And the three interior angles ABC, BCA, and CAB are equal to two right angles.

Through the point c draw the straight line, (I. 31) CE, parallel to the side BA.

B

Fig. 32. No. 1.

E

c

D

Because CE is parallel to B A, and AC meets them, the angle ACE is equal (I. 29) to the alternate angle BA C. Again, because UE is parallel to AB and BD falls upon them, the exterior angle ECD is equal (I. 29) to the interior and opposite angle AB C. But the angle

A CE was shown to be equal to the angle B AC. Therefore the whole exterior angle A CD is equal (Ax. 2) to the two interior and opposite angles C A B and A B C. To each of these equals, add the angle ACB; the two angles A CD and ACB are therefore equal (Ax. 2) to the three angles CAB, A B C and ACE. But the two angles ACD and ACB are equal (I. 13) to two right angles. Therefore also the three angles CA B, A B C and ACB are equal (Ax. 1) to two right angles. Wherefore, if a side of any triangle be produced, etc. Q. E. D.

Corollary 1.- All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Let ABCDE, fig. 32, No. 2, be any rectilineal figure. All the interior angles ABC, B C D, etc., together with four right angles, are equal to twice as many right Fig. 32, No. 2. angles as the figure has sides.

Divide the rectilineal figure ABCDE into as many triangles as the figure has sides, by drawing straight lines from a point F, within the figure to each of its angles. Because the three interior angles of a triangle are equal (I. 32) to two right angles, and there are as many triangles in the figure as it has sides, all the angles

D

EXERCISE II. TO PROPOSITION XXXII.

123

equal parts. Let A B, fig. g, be the given straight line. It is required to trisect it; that is, to divide it into three equal parts.

of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz. ABC, BC D, etc., To trisect a given finite straight line; that is, to divide it into three together with the angles at the point F. which are equal (I 15. Cor. 2) to four right angles. Therefore all the angles of these triangles are equal (4x. 1) to the interior angles of the figure together with four right angles. But it has been proved that all the angles of these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Let fig. 32, No. 3, represent any rectilineal figure.

D

A

Fig. 32. No. 3.

B

Because the interior angle ABC, and its adjacent exterior angle ABD are (I. 13) together equal to two right angles. Therefore all the interior angles together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with all the exterior angles are equal (Ax. 1) to all the interior angles and four right angles. Take from these equals all the interior angles; therefore all the exterior angles of the figure are equal (Ax. 3) to four right angles.

Corollary 1.-If two angles of a triangle be given, the third is given; for it is the difference between their sum and two right angles.

Corollary 2.-If two angles of one triangle be equal to two angies of another triangle, the third angle of the one is equal to the third angle of the other.

Corollary 3.-Every angle of an equilateral triangle is equal to one-third of two right angles, or two-thirds of a right angle. Hence, a right angle can be trisected.

Corollary 4.-If one angle of a triangle be a right angle, the sum of the other two is a right angle.

Corollary 5.-If one angle of a triangle be equal to the sum of the other two, it is a right angle.

Corollary 6.-If one angle of a triangle be greater than the sum of the other two, it is obtuse; and if less, acute. Corollary 7.-In every isosceles right-angled triangle, each of the acute angles is equal to half a right angle.

Corollary 8.-All the interior angles of every quadrilateral figure are together equal to four right angles. This is only a particular case of Euclid's Cor. 1: but it is very necessary to be remembered.

EXERCISE I. TO PROPOSITION XXXII.

If a straight line be drawn from one of the angles of a triangle,
making the exterior angle equal to the two interior and opposite
angles, it is in the same straight line with the adjacent side.
Let CD, fig. f, be a straight line drawn from the angle BCA
of the triangle A B C, making the exterior angle A C D equal to
the two interior angles CBA and BAC;
then CD is in the same straight line
with BC.

B

Fig.f.

A

Because the angle ACD is equal (Hyp.) to the two angles CBA and BAC, add to each of these equals the angle BOA. Then the two angles ACD and B CA, are equal to the three angles BCA, CAB and ABC. But these three angles (Prop. 32) are equal to two right angles. Therefore the angles A CB and ACD are also equal to two right angles. Therefore the straight line CD (Prop. 14) is in the same Q. E. D. straight line with B C.

This exercise was solved by Q. PRINGLE (Glasgow), who added other two modes of solution, equally correct; it was also solved by D. H. (Driffield): T. BocoCK (Great Warley); E. J. BREMNER (Carlisle); C. L. HAD FIELD (Bolton-le Moors); J. H. EASTWOOD (Middleton); and others.

A

G

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Through the point A draw the straight line A c, making any From CD cut angle with AB. Produce A c indefinitely to D. Join FB. Through off c E and E F each, of them equal to a c. the points c and E, draw CG and EH parallel to F B and meeting A B in G and H (Prop 31). Then A B is divided into three equal parts in the points G and H. Through the points c and. F, draw CI and E к, parallel (Prop. 31) to A B, and meeting FB in the points I and K; and let c1 meet EH in L.

Join G L.

Because AG, CL and EK are parallel (Const.), the angles FEK, ECL, and CAG are equal (Prop. 29). And because CG, EL, and FK are parallel (Const.), and A E falls upon them, the angles ACG, CEL and EFK are equal (Prop. 29). And the sides a C, CE and EF of the triangles ACG, ECL and EFK are equal (Const.); therefore, (Prop. 26) the sides AG, CL and Ex are equal. But because CL is parallel to G H, and G L meets them, the angles CLG and LG H are equal (Prop. 29); and because co is parallel to LH, and GL meets them, the angles CGL and HLG are equal (Prop. 29); and GL is common to the two triangles GCL and G HL; therefore c L is equal (Prop. 26) to a H. But CL was shown to be equal to A G. Therefore G H is equal (Ax. 1) to a G. In like manner, by joining L K and H 1, it may be shown that E K is equal to L1, and that LI is equal to H B. But E K was shown to be equal to a G. Therefore H B is equal to A G. But GH is also equal to AG. Therefore, the three straight lines AG, GH, and HB are equal to one another. Wherefore the straight line A B has been trisected in the points Q. E. F. G and H. Note. By this problem a straight line may be divided into any number of equal parts.

EXERCISE III. TO PROPOSITION XXXII. † Any angle of a triangle is right, acute, or obtuse, according as the straight line drawn from its vertex bisecting the opposite side is equal to, greater than, or less than half that side.

In fig. h, let A в be a triangle, and let a D be drawn from A, bisecting B C in D. Fig. h.

No. 1.

No. 2.

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No. 3.
B

A

This exercise was solved by Q. PRINGLE (Glasgow), as above, and he has the merit of generalising it, so as to make it applicable to the division of a straight line into any number of equal parts; a problem which Dr. Thomson postpones till he comes to Prop. xxxiv. It was also solved by WARIN (East Dereham); E. J. BREMNER (Carlisle); J. H. EASTWOOD (Middleton); and others.

This exercise was solved as above by Q. PRINGLE (Glasgow); J. Bocock (Great Warley); C. L. HADFIELD (Bolton-le Moors); H. J. PUGH (Longsight); E. J. BREMNER (Carlisle); D.`H. (Driffield); J. H. EASTWOOD (Middleton); and others.

1st. Let AD be equal to DC No. 1. a right angle.

Then the angle B A C is Because A D is equal to DC, the angle D A C is equal (Prop. 5) to the angle D C A. For a like reason, the angle DAB is equal to the angle DBA. Therefore, the angle BAC is equal to the angles ABC and A CB (Ax. 2). But the three angles BAC, ACB and ABC, are equal (Prop. 32) to two right angles. Therefore the angle BAC (Cor. 5) is a right angle.

2nd. Let AD be greater than DC No. 2. Then the angle

BAC is acute.

Because DA is greater than D C, the angle DA c is less (Prop. 18) than the angle DCA. For a like reason, the angle D A B is less than the angle DBA. Therefore the angle BAC is less then the two angles A B C and a C B. Therefore the angle B A C (Cor. 6) is acute. 3rd. Let A D be less than D C No. 3. Then the angle BAC is obtuse. Because DA is less than D C, the angle DA c is greater than the angle DCA (Prop. 18). For a like reason, the angle D A B is greater than the angle DBA. Therefore the angle BAC is greater than the angles ABC and AC B. Therefore the angle BAC (Cor. 6) is obtuse. Wherefore, any angle of a triangle is right, acute, or obtuse, etc. Q. E. D.

EXERCISE IV. TO PROPOSITION XXXII.*

The straight line drawn from the vertex of any angle of a triangle bisecting the opposite side, is equal to, greater than, or less than half that side, according as the angle is right, acute, or obtuse. In fig. h, let A B c be a triangle, and let ▲ D be drawn from a bisecting B C in D.

1st. Let the angle B A C No. 1 be a right angle. Then A D is equal to D C. For, if not, then A D is either greater or less than D c; and, according as A D is greater or less than D C, 80 is the angle B A c acute or obtuse, by the preceding exercise; but it is neither acute nor obtuse by hypothesis; therefore A D is neither greater nor less than D C-that is, A D is equal

to D C.

2nd. Let the angle B A C No. 2 be acute. Then A D is greater than D c. For, if not, then AD is either equal to, or less than, DC; and, according as A D is equal to, or less than, DC, so is the angle B A C right or obtuse, by the preceding exercise; but it is neither right nor obtuse by hypothesis; therefore A D is neither equal to, nor less than, D C-that is, A D is greater than D C.

3rd. Let the angle B A C No. 3 be obtuse. Then, it may be shown, as in the preceding cases, that A c is less than a D. Wherefore the straight line drawn, etc. Q. E. D.

EXERCISE V. TO PROPOSITION XXXII.†

If the sides of an equilateral and equiangular pentagon (or fivesided figure) be produced till they meet, the angles formed at the points of meeting are together equal to two right angles.

| sides A B, BC, CD, DE and EA, both ways till they meet in the points F, G, H, K and L. Then all the angles at these points are together equal to two right angles.

Because the angle FA E is the exterior angle of the triangle LA H, it is equal to the two interior and opposite angles at H and 1 (Prop. 32); and for a like reason, the angle A EP is equal to the angles at G and K; therefore (Ax. 2) the two angles FAE and A E F are equal to the four angles at H, L, G and K. To these equals add the angle at F; and the three angles FA E, EA F and A FE are equal to the five angles at the points F, G, H, K and L. But the three angles FE A, EAF and AFE are equal to two right angles (Prop. 32); therefore (Ax. 1) the five angles at the points F, G, H, K and L, are equal to two right angles. Wherefore, if the sides of an equilateral, etc. Q. E. D.

EXERCISE VI. TO PROPOSITION XXXII.

If the sides of an equilateral and equiangular hexagon (or six-sided figure) be produced till they meet, the angles formed at the points of meeting are together equal to four right angles.

In fig. k, let A B C DEF be a regular hexagon. Produce the sides AB, BD, DC, CE and EF, both ways till they meet in the points G, H, K, L, M and N. Then all the angles at these points are equal to four right angles.

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ANSWERS TO CORRESPONDENTS.

R. 8. T. (Leeds): Pronounce Latin in general like English. Long a is pronounced as in fate. Final is in the dative and ablative plural is pronounced as the English word ice; but many adopt the short sound, as in hiss. Sim and sis are pronounced like him, hiss; final am, as, and em, like

In fig. i, let ABCDE be a regular pentagon. Produce the the English words am, as, and them. The apparent discrepancy between

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Walker and Smith arises from the fact that the former denotes accent, the latter quantity. Walker should be followed in pronunciation.

D. Ú. N. (Slough): Pronounce o, h. To, as hee, he, toe, nv as seen, and wv as tone. It is unnecessary to learn the English-Greek exercises. The exercises are well done-particularly the English-Greek. Instead of "The king he is a general," you should have said "The king himself," etc.

H. T. S. H. (Oxford): We fear that his scheme for an air pump, however ingenious, would not answer; we strongly doubt the applicability of glass pistons. OMEGA: We must try and insert the biography he wants in the P. E.-J. Mc BAIN (Fetter-lane): The Cotton Account is not debited for the amount of cash received, but for the amount due by the merchant for the cotton purchased on credit or otherwise; and the same account is credited by the amount due for the cotton sold. The reason why the amounts due are entered, less the discounts, is because the latter are small and customary, and not worth opening a separate account for; besides, the sums, less the discounts, are the actual sums for which the cotton was purchased or sold, and are, after all, the proper sums to be entered in the Ledger. The incidental expenses arising from the sending of the cotton to a distance are, in like manner, added to the cost of the article, as constituting the real amount due for it, or the actual sums for which the cotton was purchased or sold. As to Petty Cash, its name implies that it has nothing to do with Discounts, Incidental Expenses attending the purchase or sale of goods, Interest, or any sums of money laid out in trade transactions; it is only applicable to small disbursements which occur in the counting-house, for stationary, postages, coals, candles, etc.-LABORE VINCO (Soho): His solutions of the Centenary Problems are very good, but they have come to us so irregularly and by piece-meal, that we could not insert their number in our list.

This exercise was solved by J. H. EASTWOOD (Middleton); T. BоCOCK (Great Warley); E. J. BREMNER (Carlisle); and WARIN (East Dereham).

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