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Now | ABCD = | EBCII, being on the same base
BC, and between the same parallels BC, AH;

I. 35

I. 35

=

and || EFGH : || EBCH, being on the same base EH, and between the same parallels EH, BG;

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1. Prove the proposition by joining AF, DG instead of BE, CH.

2. Divide a given ||m into two equal ||ms

3. In how many ways may this be done?

4. Of two ms which are between the same parallels, that is the greater which stands on the greater base.

5. State and prove a converse of the last deduction.

6. Equal ||ms situated between the same parallels have equal bases.

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Triangles on the same base and between the same parallels are equal in area.

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Let ABC, DBC be triangles on the same base BC, and between the same parallels AD, BC:

it is required to prove ▲ ABC =

=

▲ DBC.

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B

Through B draw BE | AC, and through C draw CF

BD;

and let them meet AD produced at E and F.

Then EBCA, DBCF are ||ms;

and between the same parallels BC, EF.

I. 31

I. Def. 33

DBCF, being on the same base BC,

I. 35

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COR.-Hence any rectilineal figure may be converted into

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Let ABCDE be any rectilineal figure:

it is required to convert it into an equivalent triangle.

Join AC, AD;

through B draw BF || AC, through E draw EG || AD, I. 31

and let them meet CD produced at F and G.

Join AF, AG.

AFG is the required triangle.

For Δ AFC = Δ ΑΒC, and Δ AGD = ▲ AED; I. 37 :. ▲ AFC + ▲ ACD + ▲ AGD = ▲ ABC + ▲ ACD + ▲ AED.

.. Δ AFG = figure ABCDE.

A DBC=

1. ABC is any triangle; DE is drawn || the base BC, and meets AB, AC at D and E; BE and CD are joined. Prove =▲ EBC, ▲ BDE = ▲ CED, and ▲ ABE = ▲ ACD. 2. ABCD is a quadrilateral having AB || CD; its diagonals AC, BD meet at O. Prove A AOD = ▲ BOC.

3. In what case would no construction be necessary for the proof

of this proposition?

4. Convert a quadrilateral into an equivalent triangle.

5. ABC is any triangle, D a point in AB; find a point E in BC produced such that ▲ DBE = ▲ ABC.

PROPOSITION 38. THEOREM.

Triangles on equal bases and between the same parallels are equal in area.

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Let ABC, DEF be triangles on equal bases BC, EF, and between the same parallels AD, BF:

it is required to prove ▲ ABC

=

Δ DEF.

Through B draw BG || AC, and through F draw FH || DE;

and let them meet AD produced at G and H.

Then GBCA, DEFH are ||ms;

I. 31

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|| DEFH, being on equal bases BC, EF,

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COR. The straight line joining any vertex of a triangle to the middle point of the opposite side bisects the triangle. Hence the theorem: If two triangles have two sides of the one respectively equal to two sides of the other and the contained anglès supplementary, the triangles are equal

in area.

1. Of two triangles which are between the same parallels, that is the greater which stands on the greater base.

2. State and prove a converse of the last deduction.

3. Two triangles are between the same parallels, and the base of the first is double the base of the second; prove the first triangle double the second.

4. The four triangles into which the diagonals divide a ||m are equal. 5. If one diagonal of a quadrilateral bisects the other diagonal, it also bisects the quadrilateral.

6. ABCD is a ||m; E is any point in AD or AD produced, and F any point in BC or BC produced; AF, DF, BE, CE are joined. Prove ▲ AFD= = A BEC.

7. ABC is any triangle; L and K are the middle points of AB and AC; BK and CL are drawn intersecting at G, and AG is joined. Prove ▲ BGC = =AAGC = ▲ AGB. 8. ABCD is a ||m; P is any point in the diagonal BD or BD produced, and PA, PC are joined. Prove ▲ PAB = A PCB, and ▲ PAD = ▲ PCD.

9. Bisect a triangle by a straight line drawn from a given point in

one of the sides.

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Equal triangles on the same side of the same base are between the same parallels.

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Let As ABC, DBC on the same side of the same base BC be equal, and let AD be joined :

it is required to prove AD || BC.

If AD is not || BC, through A draw AE || BC, meeting BD, or BD produced, at E, and join EC.

I. 31

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which is impossible, since the one is a part of the other. ... AD is || BC.

I. 37

Hyp.

1. The straight line joining the middle points of two sides of a triangle is the third side, and half of it.

=

2. Hence prove that the straight line

joining the middle point of the hypotenuse of a right-angled triangle to the opposite half the hypotenuse.

vertex =

=

3. The middle points of the sides of any quadrilateral are the vertices of a m, whose perimeter the sum of the diagonals of the quadrilateral. When will this m be a rectangle, a rhombus, a square?

4. If two equal triangles be on the same base, but on opposite sides of it, the straight line which joins their vertices will be bisected by the base.

5. Use the first deduction to solve I. 31.

6. In the figure to I. 16, prove AF || BC.

7. If a quadrilateral be bisected by each of its diagonals, it is a ||TM.

8. Divide a given triangle into four triangles which shall be equal in every respect.

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