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hypotenuse, and the other bisecting it, they shall contain an angle equal to the difference of the two acute angles of the triangle.

Let the A, of the A BAC be a right; let

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AE be drawn to the bisection E, of the hypotenuse BC, and let AD be drawn perpendicular to BC: The 4 EAD=4C-4 B.

For (S. 29. 1. and hyp.) EA= EB; .. the EAB=2EBA: Again, since (hyp.) the two BAC, CBA, of the A BAC, are equal to the two ■ BDA, ABD, of the ▲ ADB, .. (S. 26. 1.) the < BAD=4 ACB; but the EAD=2 BAD

BAE; .. the

EAD=4C-4 B.

71. PROBLEM.

PROP. LIV.

To bisect a parallelogram by a

straight line drawn through a given point in one of its sides.

Let ABCD be a□, and E a given point in one of its sides: It is required to bisect the

by a straight line drawn through E.

ABCD,

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From AD, the side opposite to BC, cut off DF=BE (E. 3. 1.); and join E, F; EF bisects the ABCD.

For, through E and F draw (E. 31. 1.) EG and FH, each parallel to AB or DC; ... AE, GH, FC are; and since EF is the diameter of the GH, .. (E. 34. 1.) the A EGF-A EHF; also, because BEFD, and that AD is parallel to BC, .. (E. 36. 1.) the AE FC; to these equals add the equal, EGF, EHF, and it is evident that the trapezium ABEF is equal to the trapezium FECD; i. e. EF bisects the ABCD.

PROP. LV.

72. THEOREM. A trapezium, which has two of its sides parallel, is the half of a rectangle between the same parallels, and having its base equal to the aggregate of the two parallel sides of the trapezium.

Let ABEF be a trapezium, having its side AF parallel to the opposite side BE; The trapezium ABEF is equal to the half of a rectangle between

F

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AF and BC, and having its base equal to AF+ BE.

For, produce BE to C, and make EC=AF; through C draw (E. 31. 1.) CD parallel to BA, and let CD meet AF produced in D; .. the figure ABCD is a ;.. (E. 34. 1.) AD= BC; and (constr.) AF=EC; .. FD=BE: It is manifest, .., (from S. 54.1.) that the trapezium ABEF is the half of the ABCD; but (E. 35. 1.) the ABCD a rectangle upon the same base BC, and between the same two parallels ; ... the trapezium ABEF= the half of a rectangle on the base BC, which (constr.) = BE + AF, and between the two parallels BE and AF.

PROP. LVI.

73. PROBLEM. Any two parallelograms having been described on two sides of a given triangle, to apply, to the remaining side, a parallelogram, which shall be equal to their aggregate.

Let the AQ and AP be on the two sides AB, AC, of the given ▲ ABC: It is required to apply

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equal to the AP together with the AQ. Produce QD and PE until they meet in F; join F, A; through C draw (E. 31. 1.) CG parallel to FA, and make, also, CG FA; complete the

BCGH: The BCGH=□AQ+□ AP. For, produce FA, so that it shall meet BC in I, and HG in K; produce, also, GC and HB, until they meet EP and DQ in L and M; .. the figures FACL, FABM are ; and, since (constr.) the

FACL, CGKI, are upon equal bases FA, CG and between the same parallels, .. (E. 36. 1.) the FACL CGKI; but (E. 35. 1.) the☐ FACL=□ AP; .. the AP,=□GI: And in the same manner, it may be proved that the AQ=□IH; .. the whole BCGH +OAQ.*

AP

* If the parallelograms AP and AQ are squares, it is easy

PROP. LVII.

74. PROBLEM. A plane rectilineal figure of any number of sides being given, to find an equal rectilineal figure, which shall have the number of its sides less, or greater, by one, than that of the given figure.

First, let ABCDE be a given rectilineal figure:

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It is required to find an equal rectilineal figure, having the number of its sides less by one, than the number of the sides of ABCDE.

of the

Let A, E, D be any three consecutive given figure ABCDE; join A, D ; through E draw (E. 31. 1.) EF parallel to AD and meeting CD, produced, in F; join A, F: The figure ABCF, which has the number of its sides less by one than ABCDE, is equal to ABCDE.

For, since the two AED, AEF, are upon

to shew that the parallelogram BG will also be a square; and thus the forty-seventh proposition of the first Book of Euclid's Elements will have been demonstrated.

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