Elements of geometry, containing books i. to vi.and portions of books xi. and xii. of Euclid, with exercises and notes, by J.H. Smith

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 angle.

Let AB be the given st. line, C the given ▲, D the given 4.

It is required to apply to AB a □ = ▲ C and having one of its LS= LD.

Make a =AC, and having one of its angles ▲ D,

1. 42.

and suppose it to be removed to such a position that one of the sides containing this angle is in the same st. line with AB, and let the be denoted by BEFG.

Produce FG to H, draw AH || to BG or EF, and join BH. Then FH meets the s AH, EF,

.. sum of 8 AHF, HFE=two rt. 48;

.. sum of 4s BHG, HFE is less than two rt. 2s;

:. HB, FE will meet if produced towards B, E.

Let them meet in K.

Through K draw KL || to EA or FH,

Post. 6.

and produce HA, GB to meet KL in the pts. L, M. Then HFKL is a □, and HK is its diagonal;

and AG, ME ares about HK,

.. complement BL= complement BF,

:: OBL= A C.

Also the BL has one of its 4s, ABM= LEBG, and

equal to 4 D.

Q.E.F.

PROPOSITION XLV. PROBLEM.

To describe a parallelogram, which shall be equal to a given rectilinear figure, and have one of its angles equal to a given angle.

Let ABCD be the given rectil, figure, and E the given 4.

Join AC.

It is required to describe a☐ = to ABCD, having one of its LS LE.

Describe a FGHK= ▲ ABC, having / FKH= L E.

I. 42.

To GH apply a □ GHML= ▲ CDA, having ▲ GHM= LE.

Then FKML is the reqd.

For :: LGHM and FKH are each ▲ E;

.. LGHM= L FKH,

I. 44.

.. sum of 4s GHM, GHK=sum of 48 FKH, GHK

=two rt. 48;

.. KHM is a st. line.

Again, HG meets the s FG, KM,

LFGH= LGHM,

I. 29.

I. 14.

.. sum of 4s FGH, LGH=sum of 48 GHM, LGH

=two rt. 48;

.. FGL is a st. line.

Then :: KF is || to HG, and HG is || to LM,

.. KF is to LM;

and KM has been shewn to be || to FL,

:. FKML is a parallelogram,

and :: FH= ^ ABC, and GM=▲ CDA,

I. 29.

I. 14.

I. 30.

In the same way a☐ may be constructed equal to a given rectil. fig. of any number of sides, and having one of its angles

equal to a given angle.

Q. E. F.

Miscellaneous Exercises.

1. If one diagonal of a quadrilateral bisect the other, it divides the quadrilateral into two equal triangles.

2. If from any point in the diagonal, or the diagonal produced, of a parallelogram, straight lines be drawn to the opposite angles, they will cut off equal triangles.

3. In a trapezium the straight line, joining the middle points of the parallel sides, bisects the trapezium.

4. The diagonals AC, BD of a parallelogram intersect in O, and P is a point within the triangle AOB; prove that the difference of the triangles APB, CPD is equal to the sum of the triangles APC, BPD.

5. If either diameter of a parallelogram be equal to a side of the figure, the other diameter shall be greater than any side of the figure.

6. If through the angles of a parallelogram four straight lines be drawn parallel to its diagonals, another parallelogram will be formed, the area of which will be double that of the original parallelogram.

7. If two triangles have two sides respectively equal and the included angles supplemental, the triangles are equal.

8. Bisect a given triangle by a straight line drawn from a given point in one of the sides.

9. If the base of a triangle ABC be produced to a point D such that BD is equal to AB, and if straight lines be drawn from A and D to E, the middle point of BC; prove that the triangle ADE is equal to the triangle ABC.

10. Prove that a pair of the diagonals of the parallelograms, which are about the diameter of any parallelogram, are parallel to each other.

PROPOSITION XLVI. PROBLEM.

To describe a square upon a given straight line.

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Let AB be the given st. line.

It is required to describe a square on AB.
From A draw ACL to AB.

In AC make AD=AB.
Through D draw DE || to AB.
Through B draw BE || to AD.

Then AE is a parallelogram,

and.. AB=ED, and AD-BE.

But AB= AD,

Ex. 1.

.. AB, BE, ED, DA are all equal;
.. AE is equilateral.

And 4 BAD is a right angle,

.. AE is a square,

and it is described on AB.

Def. xxx.

Q. E. F.

Shew how to construct a rectangle whose sides are equal to two given straight lines.

Ex. 2. Shew that the squares on equal straight lines are equal.

Ex. 3. Shew that equal squares must be on equal straight lines.

NOTE. The theorems in Ex. 2 and 3 are assumed by Euclid in the proof of Prop. XLVIII. See also Prop. A. of Book II.,

p. 77.

PROPOSITION XLVII. THEOREM.

In any right-angled triangle the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle.