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For since ADBC is a □, AD—CB (E. 34, 1.) and (E. 29. 1.) the EAD of the A AED, EBC, of the ▲ BEC, and the EDA ECB; (E. 26. 1.) AE= EB, and DE EC.


58. THEOREM. If in two opposite sides of a parallelogram two points be assumed, one in each of those sides, equidistant from two opposite angles of the figure, and if two other points be likewise assumed, in the two other opposite sides, equidistant from the same two angles, the figure, contained by the straight lines joining the four points so assumed, shall be a parallelogram.

In the opposite sides AD, BC of the ABCD,

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let the points E and G be taken equidistant from the opposite A and C ; let also, the points F and H be taken, in the other two opposite sides, AB and DC, equidistant from A and C; and let E, F, and F, G, and G, H, and H, E, be joined: The figure EFGH is a parallelogram.


For since (hyp.) AE=CG, and AF CH, and AC,... (E. 4. 1.) FE➡

that (E. 34. 1.) the

GH: Again since (E. 34. 1.) AB=DC, and ADBC, and that (hyp.) of AD the part AE is equal to the part CG of BC, and of AB the part AF is equal to the part CH of DC, .'. ED=GB, and DH BF; also (E. 34. 1.) the EDH= (E. 4. 1.) EHFG; and it has been

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< FBG; proved that EF a parallelogram.

HG; ... (S. 18. 1.) EFGH is


59. THEOREM. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point.

Let ABCD be a given, and let EFGH be

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anywhatever, inscribed in ABCD: The diameters of ABCD and of EFGH cut one another in the same point.

For draw AC a diameter of ABCD, and FH a diameter of EFGH; let AC and FH cut one another in K; and let CB, produced, meet EF, produced, in L: Then, since AE is parallel to BC, and EF parallel to HG, the CGH (E. 29. 1.)

GLE; and the GLE 2 LEA; = < LEA; .. the < CGH AEF; also (hyp. and E. 34. 1.) the A=C, and the side FE the opposite side GH, of the EFG; ... (E. 26. 1.) CH=AF: Again, since the side AF of the ▲ AKF= the side CH of the ▲ CKH, and that (E. 29. 1.) the

KAF, KFA are equal to the KCH, KHC, .`. (E. 26. 1.) AK=KC, and FK=KH; i. e. K is the bisection of the diameters AC, FH; .. (S. 42. 1.) all the diameters cut one another in the point K.

60. COR. From the demonstration it is manifest, that the angle contained by any two given straight lines, is equal to the angle contained by two other straight lines, that are parallel to the two given straight lines, each to each.



The diameters of an equilateral four-sided plane rectilineal figure bisect one another at right angles.

Let AB and DC be the diameters of the equilateral four-sided figure ACBD, cutting one another in E: AB and DC bisect one another in E, at right angles.

For, since (hyp.) ACBD is equilateral, it is (S. 18. 1.) a □; and .·. (S. 42. 1.) the diameters bisect one another in E: Again, because DE CE, and EA is common to the two


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AEC, and that (hyp.) ADAC, .‘. (E. 8. 1.) AED AEC; i. e. (E. 10. def. 1.) each

of the

AED, AEC is a right 4; .'. (E. 15. 1.) each of the DEB, CEB, is, also, a right .


62. THEOREM. The diameters of a rectangle are equal to one another.

Let AC, and BD be the diameters of the rectangle ABCD: Then ACBD.

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For, since (hyp.) the opposite of the figure are equal, each being a right ▲, .'. (S. 26. 1.) the

figure ABCD is a □ ; ... •. (E. 34.-1.) AD—BC; and AB is common to the two ABC, BAD,

and the BD.

ABC BAD; ... (E. 4. 1.) AC=


63. PROBLEM. To inscribe a square in a given equilateral four-sided figure.

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figure It is required to inscribe a square in ABCD.

Join A, C, and B, D, and let AC and BD cut one another in X; bisect (E. 9. 1. and E. 15. 1.) the AXB and CXD, by the straight line EG, and the BHC, AXD, by the straight line FH; and join E, F, and F, G, and G, H, and H, E: The inscribed figure EFGH is a square.

For, since the figure ABCD is (hyp.) equilateral, AC and BD (S. 45. 1.) bisect one another at right ;.. (constr.) each of the EXA, AXH, HXD, DXG, GXC, CXF, FXB, BXE, is half a right .. the EXH, HXG, GXF, FXE are right : Again, because ABCD is equilateral, it

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