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PROPOSITION XXXIV. THEOREM.

The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it.

Let ABDC be a 0, and BC a diagonal of the O.

Then must AB=DC and AC=DB, and

_ BAC= 2 CDB, and ABD= . ACD and

A ABC= ADCB. For :: AB is || to CD, and BC meets them, ... ABC=alternate 2 DCB ;

I. 29. and :: AC is ll to BD, and BC meets them, ..LACB=alternate . DBC.

I. 29. Then in As ABC, DCB,

:: LABC= _ DCB, and ACB= 2 DBC, and BC is common, a side adjacent to the equal Zs in each ; .. AB=DC, and AC=DB, and 2 BAC= _ CDB,

and A ABC= ADCB. Also :: 2 ABC= _ DCB, and < DBC= 1 ACB, .: Z8 ABC, DBC together= 4s DCB, ACB together, L ABD= L ACD.

Q. E. D. Ex. 1. Shew that the diagonals of a parallelogram bisect each other.

Ex. 2. Shew that the diagonals of a rectangle are equal.

I. B.

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PROPOSITION XXXV. THEOREM. Parallelograms on the same base and between the same parallels are equal.

D E

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Let the Os ABCD, EBCF be on the same base BC and between the same ||s AF, BC.

Then must O ABCD=O EBCF.
CASE I. If AD, EF have no point common to both,
Then in the As FDC, EAB,
:: extr. - FDC=intr. _ EAB,

I. 29. and intr. 2 DFC=extr. 2 AEB,

I. 29. and DC=AB,

I. 34.

.:. Δ FDC=Δ ΕΑΒ.

I. 26.

Now O ABCD with 4 FDC=figure ABCF; and D EBCF with A E AB=figure ABCF ; .:. O ABCD with a FDC=D EBCF with a EAB;

:. | ABCD=G EBCF.

CASE II. If the sides AD, EF overlap one another,

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Case III. If the sides opposite to BC be terminated in the same point D,

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the same method of proof is applicable,

but it is easier to reason thus : Each of the Ds is double of a BDC;

I. 34. .. O ABCD=O DBCF.

Q. E, D. PROPOSITION XXXVI. THEOREM. Parallelograms on equal bases, and between the parallels, are equal to one another.

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B

Let the Os ABCD, EFGH be on equal bases BC, FG, and between the same Is AH, BG.

Then must D ABCD=O EFGH.

Join BE, CH. Then .: BC=FG,

Нур. and EH=FG ;

I, 34. .: BC=EH; and BC is | to EH.

Нур. :. EB is || to CH;

I. 33. :: EBCH is a parallelogram. Now O EBCH=ABCD,

I. 35. ::: they are on the same base BC and between the same lls; and D EBCH=DEFGH,

I. 35. : they are on the same base EH and between the same s;

.:D ABCD=D EFGH.

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Q. E. D.

PROPOSITION XXXVII. THEOREM.

Triangles upon the same base, and between the same parallels, are equal to one another.

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Let As ABC, DBC be on the same base BC and between the same || s AD, BC.

Then must A ABC= ADBC.

From B draw BE | to CA to meet DA produced in E.
From C draw CF || to BD to meet AD produced in F.
Then EBCA and FCBD are parallelograms,
and O EBCA=D FCBD,

I. 35. :: they are on the same base and between the same ||s. Now A ABC' is half of D EBCA,

I. 34. and a DBC is half of O FCBD;

I. 31. .: A ABC= A DBC.

Ax. 7.

Q. E. D. Ex. 1. If P be a point in a side AB of a parallelogram ABCD, and PC, PD be joined, the triangles PAD, PBC are together equal to the triangle PDC.

Ex. 2. If A, B be points in one, and C, D points in another of two parallel straight lines, and the lines AD, BC intersect in E, then the triangles AEC, BED are equal.

PROPOSITION XXXVIII. THEOREM.

Triangles upon equal bases, and between the same parallels, are equal to one another.

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Let As ABC, DEF be on equal bases, BC, EF, and between the same || s BF, AD.

Then must A ABC= A DEF.

From B draw BG || to CA to meet DA produced in G. From F draw FH || to ED to meet AD produced in H. Then CG and EH are parallelograms, and they are equal,

they are on equal bases BC, EF, and between the same lls BF, GH.

I. 36.
Now A ABC is half of O CG,
and A DEF is half of O EH;
..A ABC=ADEF.

Ax. 7.
Q. E. D.

Ex. 1. Shew that a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts.

Ex. 2. In the equal sides AB, AC of an isosceles triangle ABC points D, E are taken such that BD=AE. Shew that the triangles CBD, ABE are equal.

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