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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.

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

It is required to apply to AB a ☐ =AC and having one of its 48= D.

Make a

Also the .. equal to 4 D.

▲ C, and having one of its angles= ▲ D, I. 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 Is AH, EF,

.. sum of 4s AHF, HFE=two rt. 4 s ;

I. 29.

.. sum of 4 s BHG, HFE is less than two rt. s; :. HB, FE will meet if produced towards B, E. Post. 6. Let them meet in K.

Through K draw KL || to EA or FH,

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,
:. BL= A C.

I. 43.

BL has one of its 4s, ABM= ▲ EBG, and

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.

G I

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D

K

H MA

B

=

Let ABCD be the given rectil. figure, and E the given 2. It is required to describe a ☐ to ABCD, having one of its s=LE.

Join AC.

Describe a FGHK= ▲ ABC, having ▲ FKH= 2 E.

I. 42.

To GH apply a □ GHML= ^ CDA, having ▲ GHM= 2 E. I. 44.

Then FKML is the reqd. For GHM and ▲ FKH are each= ▲ E ; .. ¿ GHM= ▲ FKH,

.. sum of 4s GHM, GHK=sum of 4s FKH, GHK
=two rt. 4S;
I. 29.
I. 14.

.. KHM is a st. line.

Again, HG meets the ||s FG, KM,

LFGHL GHM,

.. sum of 8 FGH, LGH=sum of 48 GHM, LGH
=two rt. 4 S;

I. 29.
I. 14.

.. 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,
..FM=whole rectil. fig. ABCD,

and FM has one of its 4 s, FKM= ▲ E. 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.

I. 30.

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 diagonal of a parallelogram be equal to a side of the figure, the other diagonal 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 parallelograin 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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Q. E. F.

Ex. 1. 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.

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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.

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D L

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Then in As ABD, FBC,

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Let ABC be a right-angled ▲, having the rt. 4 BAC. Then must sq. on BC= sum of sqq. on BA, AC. On BC, CA, AB descr. the sqq. BDEC, CKHA, AGFB. Through A draw AL || to BD or CE, and join AD, FC. BAC and 4 BAG are both rt. 4s,. . CAG is a st. line;

Then

and BAC and 4 CAH are both rt. 4s;
...BAH is a st. line.
Now DBC= ▲ FBA, each being a rt. 4,
adding to each 4 ABC, we have
L ABDL FBC.

I. 14.

I. 14.

Ax. 2.

·.· AB=FB, and BD=BC, and 2 ABD= 2 FBC, .. A ABD=&FBC.

I. 4.

Now BL is double of ▲ ABD, on same base BD and between same ||s AL, BD,

I. 41. and sq. BG is double of ▲ FBC, on same base FB and between same s FB, GC;

I. 41.

.. BL=sq. BG.

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