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angles (I. 29); wherefore the angles BHF and HFE, are less than two right angles. But straight lines, which, with another straight line, make the interior angles upon the same side less than two right angles, do meet if produced far enough (I. 29, Cor.);
therefore HB and FE shall meet, if produced ; let them meet in K, and through K draw KL parallel to EA or FH,
and produce HA and GB to the points L and M. (Dem.) Then HLKF is a parallelogram, of which the diagonal is HK;
and AG, ME, are the parallelograms about HK; and LB, BF, are the complements; therefore LB is equal to BF (I. 43).
But BF is equal to the triangle. C; wherefore (Ax. 1) LB is equal to the triangle C; and because the angle GBE is equal to the angle ABM (I. 15), and likewise to the angle D; the angle ABM is equal to the angle D; therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D.
PROPOSITION XLV. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Given the rectilineal figure ABCD, and the rectilineal angle E.
It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.
(Const.) Join DB, and describe (I. 42) the parallelogram FÌ equal to the triangle ABD, and having the angle HKF equal to the angle E; and (I. 44) to the straight line GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. (Dem.) And because the angle E is equal to each of the angles FKH and GHM, angle FKH is equal to GHM;
add to each of these the angle KHG; therefore the angles FKH and KHG are equal to the angles KHG and GHM; but FKH and KHG are equal to two right angles (I. 29); therefore
M ... also KHG and GHM are equal to two right angles; therefore KH is in the same straight line (I. 14) with HM. And because the straight line HG meets the parallels KM and FG, the alternate angles MHG and HGF are equal; add to each of these the angle HGL; therefore the angles MHG and HGL are equal to the angles HGF and HGL; but the angles MHG and HGL are together equal to two right angles; wherefore also the angles HGF and HGL are equal to two right angles, and FG is therefore in the same straight line (I. 14) with GL. And because KF is parallel to HG, and HG parallel to ML, KF is parallel to ML (I. 30); but KM and FL are parallels; wherefore KFLM is a parallelogram. And because the triangle ABD is equal to the parallelogram FH, and the triangle DBC to the parallelogram GM, the whole rectilineal figure ABCD is equal to the whole parallelogram FKML; therefore the parallelogram FKML has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. COR.-From this it is manifest how to a given straight line to
apply a parallelogram which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure-namely, by applying to the given straight line a parallelogram equal to the first triangle ABD (I. 44), and having an angle equal to the given angle.
PROPOSITION B. PROBLEM. To describe a triangle that shall be equal to a given rectilineal figure.
Given the figure ABCDE, it is required to construct a triangle equal to it. (Const.) Produce the side AB both ways towards F and G;
join D, B, and from C draw CG parallel to DB, and join D, G. Also, join A, D, and through E draw EF parallel to AD, and join D, F. FDG E is the triangle required.
(Dem.) For, since DB and GC are parallel, the triangles DGB and DCB are equal (I. 37). For a similar rea- F
G son, the triangles AED and AFD are equal; therefore the triangles AFD and BGD are together equal to AED and BCD; and adding ADB to these equals,
the whole triangle FDG is equal to the given figure ABCDE.
Schol.-By means of this problem any crooked boundary may be rectified; that is, a straight line may be found that will cut off the same space on each side of it that the crooked boundary does. By means of the parallel ruler, this straight boundary may be easily found. The problem is of great utility in landsurveying.
PROPOSITION XLVI. PROBLEM.
Given the straight line AB; it is required to describe a square upon AB.
(Const.) From the point A draw (I. 11) AC at right angles to AB, and make (I. 3) AD equal to AB, and (I. 31) through the point. D draw DE parallel to AB, and through B draw BE parallel to AD; (Dem.) then ADEB is a parallelogram;
whence AB is equal to DE (I. 34), and AD to BE. But BA is equal to AD; therefore the four straight lines BA, AD, DE, and EB are equal to one another, and the parallelogram ADEB is equilateral. It is cl likewise rectangular; for the straight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equal to two right angles (I. 29). But BAD is a right angle; therefore also ADE is a right angle. Now, the opposite angles of parallelograms are equal (1. 34); therefore each of the opposite angles ABE and BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. CoR.-Hence every parallelogram that has one right angle has
all its angles right angles. ; .
PROPOSITION XLVII. THEOREM. In any right-angled triangle, the square which is described upon the side opposite to the right angle, is equal to the squares described upon the sides which contain the right angle.
Given a right-angled triangle ABC, having the right angle BAC; to prove that the square described upon the side BC is equal to the sum of the squares described upon BA and AC.
(Const.) On BC describe (I. 46) the square BDEC, and on BA and AC, the squares GB and HC; and (I. 31) through A draw AL parallel to BD or CE, and join AD, FC.
(Dem.) Then, because each of the angles BAC and BAG is a right angle (Def. 40), the two straight lines AC and AG are in the same straight line (I. 14); for the same reason, : D
E AB and AH are in the same straight line; and because the angle DBC is equal to the angle FBA,
each of them being a right angle, add to each the angle ABC, and the whole angle ABD is equal to the whole angle FBC (Ax. 2); and, because the two sides AB and BD are equal to the two FB and BC, each to each, and the angle ABD equal to the angle FBC, therefore (I. 4) the base A.D is equal to the
base FC, and the triangle ABD to the triangle FBC. Now, the parallelogram BL is double of the triangle ABD (I. 41), because they are upon the same base BD, and between the same parallels BD, AL;. and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal to one another (Ax. 6); therefore the parallelogram BL is equal to the square GB. And, in the same manner, by joining AE and BK, it is demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB and HC;
and the square BDEC is described upon the straight line BC, and the squares GB and HC upon BA and AC; wherefore the square upon the side BC is equal to the two squares upon the sides BA and AC.
COR.--The square on either of the sides of a right-angled
i triangle, is equal to the difference between the squares on ...the hypotenuse and the other side.
For BC2 = AB? + AC?, and taking AC2 from both these equals, there remains BC? – AC2 = AB?; or taking AB? from both, then BC2 – ABP = AC?
Or calling the hypotenuse, the base, and the perpendicular, H, B, and P, respectively, H’ = BP + P2. Therefore H? — B2 =P; and H2 – P2 = B?.
Otherwise : A very simple demonstration of this important proposition may be given as follows: Describe the squares GB, BE, and CH on the three sides AB, BC, and CA, and join GH, FA, and AK;
describe the triangle LED, having the side LE equal to AB, and LD equal to AC, and join AL, then since DE is equal to BC (I. 34), the triangle F LDE is equal to the triangle ACB in every respect (I. 8). The angles BAČ and CAH being each right angles, BA and AH are in the same straight line (I. 14); for a like reason, CA and AG are in the same straight line. Again, FAK is a straight line, for GA and AF are equal to BA and AF, and the base GF is equal to the base BF, therefore the angle GAF is equal to the angle BAF (I. 8); but GAB is a right angle, therefore GAF and BAF are each equal to half a right angle; for a like reason, each of the angles HAK and CAK is half a right angle, and hence the two angles BAF and BAK are together equal to two right, and therefore FA and AK are in the same straight line (I. 14). It is also evident that the triangles GAH and ELD are each equal in every respect to the triangle BAC. Since the triangle FBA is equal to the triangle FGĀ, and the triangle BAC is equal to the triangle GAH, and also the triangle CAK is equal to the triangle HAK, the whole figure FBCK is equal to the whole figure FGHK; but the figure FBCK is equal to each of the figures ABDL and LECA, for if the figure FBCK be conceived to be turned round the point C through a right angle, so that CK may coincide with CA, the point K will coincide with A, because CK is equal to CA, also BC will coincide with CE, and the point B with the point E; and since the angle CBF is evidently equal to the angle CEL, the line BF will coincide with EL, and they are equal by construction since each is equal to AB; hence the point F will coincide with L, and the line FK with the line LA; hence the figure FBCK coincides with the figure LECA, and is therefore equal to it. In the same manner, if the figure FBCK be conceived to be turned round the point B through a right angle,
it may be shewn that it will coincide with the figure ABDL, and be equal to it; hence the two figures ABDLEC and FBCKHG are equal, being each double of the figure FBCK ;
but if from the first the triangles BAC and ELD be taken, there remains the square BE, and if from the second there be taken the two equal triangles BAC and GAH,
there remains the two squares GB and HC; but BE is the square on BC, and GB and HC are the squares on AB and AC, therefore the square on BC is equal to the squares on BA and AC.
PROPOSITION XLVIII. THEOREM. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle.
Given the square described upon BC, one of the sides of the triangle ABC, equal to the squares upon the other sides BA and AC; to prove that the angle BAC is a right angle.
(Const.) From the point A draw (I.:11) AD at right angles to AC, and make AD equal 5 to BA, and join DC. (Dem.) Then, because DA is equal to AB, the square on DA is equal to the square on AB. To each of these add the square on AC; therefore the squares on DA and AC are equal to the squares on BA and AC. But the square on