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PROPOSITION XLIV. PROBLEM.
To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Let AB be the given st. line, C the given ▲, and D the given 2.
It is required to apply to AB a = A C, and having one of its ▲ s = D.
so that BE be in the same st. line with AB;
produce FG to H, draw AH || BG or EF, (i. 31) and join HB.
Dem. Then, HF falls on the ||s AH, EF, .. Ls AHF,
HFE are together two rt. 4s;
..s BHF, HFE are less than two rt.
(i. 29) s:
but st. lines which with another st. line make the two int. s on the same side less than two rt. s, will meet on that side, if produced far enough:
.. HB, FE shall meet, if produced;
let them meet in K, through K draw KL || EA or FH, and produce HA, GB to meet KL at L, M.
Then KLHF is a , of which the diameter is HK;
and AG, ME are the 7s about HK, also
.. compl. LB = compl. BF ;
but BF AC; (constr.).. LB = ^ C.
And . 4. GBE = ▲ ABM, (i. 15) and also = ▲ D ;.
.. 2 ABM = 4. D.
.. to the st. line AB the
LB is applied as required.
PROPOSITION XLV. PROBLEM.
To describe a parallelogram equal to a given rectilineal figure, having an angle equal to a given rectilineal angle. Let ABCD be the given rectil. figure, and E the given 2.
It is required to describe a = to ABCD, having an E.
.. Ls FKH, KHG = 4s KHG, GHM,
but FKH, KHG together two rt. Zs;
.. KHG, GHM together two rt. Zs;
and at H in GH the two st. lines KH, HM, on the opposite sides of it, make the adj. 4s KHG, GHM = two rt. 4s, .. HK is in the same st. line with HM.
And. HG meets the s KM, FG, .. ▲ MHG
add to each of theses the
.. Ls MHG, HGL
s HGF, HGL;
KF is ||
buts MHG, HGL= two rt. Zs;
.. also HGF, HGL = two rt. Zs, and
and FL has been proved || KM, .. FKML is a ☐ ;
GM = A DBC:
HF = A ABD, and
.. the whole
KFLM has been described as re
Q. E. F.
Cor. From this it is manifest how to a given straight line to apply a parallelogram which shall be equal to a given rectilineal figure, and shall have an angle equal to a given rectilineal angle; namely by applying to the given line a parallelogram equal to the first triangle (i. 44), and having an angle equal to the given angle; and so on.
[This appears to be a difficult prop. for beginners, as some step in the demonstration is often omitted. The main point is to show clearly that FGL, KHM are straight lines by means of props. 14 and 29. When the figure FM has thus been proved to be a parallelogram, the rest easily follows.]
PROPOSITION XLVI. PROBLEM.
To describe a square on a given straight line. Let AB be the given st. line.
It is required to describe a sq. on AB.
Constr. From A draw AC AB; (i. 11) make AD = AB, through D draw DE || AB, and through B, BE || AD;
Then shall ADEB be a sq.
Dem. Since AE is a, (constr.) .. AB = DE _and
Again, since AD meets the ||s AB, DE ../s BAD,
ADE are together
two rt. s;
but BAD is a rt. ; (constr.) .. also ADE is a rt. .
Now the opp. ▲s of
s are =; (i. 34) .. each of
the opp. s ABE, BED is a rt. ;
.. AE is rectangular, and it has been proved equilateral; .. it is a sq. (def. 30) and it is described on the given st. line
Q. E. F.
that has one rt. has all its s