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.

Let ABC be a rt.-angled A, having the rt. 4 BAC.
Then the sq. on BC= the sqs. on BA, AC.

Constr. On BC describe the sq. BDEC, and on BA, AC

the sqs. GB, HC;

(i. 46)

through A draw AL || BD or CE; (i. 31) and join AD, FC.

Dem. BAC is a rt. (hyp.), and BAG is a rt. L, (def. 30) the two st. lines CA, AG on the opp. sides of AB, make with it at A the adj. s two rt. Zs; .. CA is in the same st. line with AG.

line.

(i. 14)

For the same reason, BA and AH are in the same st.

And ... AB, BD = FB, BC, each to each, and the included ABD = the included FBC,

.. A ABD = A FBC.

(i. 4)

Now C BL is double of ▲ ABD, ... they are on the same base BD and between the same ||s BD, AL;

(i. 41)

also the sq. GB is double of ▲ FBC, they are on the same base FB and between the same ||s FB, GC.

(i. 41) But the doubles of equals are equal to one another;

(ax. 6)

Similarly, by joining AE, BK, it can be proved that

..the whole sq. BDEC = the two sqs. GB, HC;

that is, the sq. on BC= the sqs. on AB, AC. Therefore, in any rt.-angled ▲, &c,

(ax. 2)

Q. E. D.

[It is essential (as in prop. 45) to show that CAG, BAH are st. lines. This is often omitted. The learner should apply this prop. to find a sq. equal to the sum of any number of given sqs., or equal to the difference of two given sqs.]

1. Four times the square on the perpendicular from an angle of an equilateral triangle on the opposite side is three times the square on one of the sides.

2. If A be the vertex of an isosceles triangle ABC, and CD be perpendicular to AB, the squares on the three sides are together equal to the square on BD, twice the square on AD, and thrice the square on CD.

3. From any point in the diameter of a semicircle two lines are drawn to the circumference, one to the middle

point of the arc, the other at right angles to the diameter. The squares on these lines are together double the square on the radius.

4. From the middle point of a side of a right-angled triangle a perpendicular is drawn to the hypotenuse : show that the difference of the squares on the segments into which it is divided is equal to the square on the

other side.

5. AB is the common hypotenuse of two right-angled triangles, ACB, ADB, and AE, BF are perpendicular to CD (produced both ways if necessary). Show that the sum of the squares on CE, CF is equal to the sum of the squares on DE, DF.

6. ABCD is a rectangle, E any point. Show that the squares on EA, EC are together equal to the squares on EB, ED.

7. If two exterior angles of a triangle be bisected, the line drawn from the point of intersection of the bisecting lines to the opposite angle of the triangle will bisect that angle.

8. Find a point in the diagonal of a square produced, from which if a straight line be drawn parallel to one of the sides, and meeting another side produced, it will form with the produced diagonal and produced side a triangle equal to the square.

9. Divide a given line into two parts such that the sum of their squares may be equal to a given square. When is this impossible?

10. On AC the hypotenuse of a right-angled isosceles triangle a square ACED is described, and BE meets AC in F. Show that the square on BE is five times the square on a side of the triangle, and that nine times the square on DF is thirteen times the square on the hypo

tenuse.

PROPOSITION XLVIII. THEOREM.

If the square described on one of the sides of a triangle, be equal to the squares described on the other two sides of it; the angle contained by these two sides is a right angle.

Let the sq. described on BC, one of the sides of A ABC, the sqs. on the other two sides AB, AC. Then shall BAC be a rt. L.

Constr. From A draw ADL AC, (i. 11), make
AD = AB, (i. 3), and join DC.

Dem.

·.· AD = AB .. sq. on AD = sq. on AB;
to each of these =s

add the sq. on AC;

And

.. sq. on DC = sq. on BC; .. DC= BC.

=

AD AB, and AC is common to As DAC, BAC;

.. AD, AC = AB, AC each to each; and the base DC has been shown the base BC; .. 4 DAC = BAC ;

(i. 8) but DAC is a rt. 4, (constr.); .. BAC is a rt. 4.

Therefore, if the sq. described on, &c. Q. E. D.

[This prop. (which is the converse of prop. 47) concludes the third section of Book i., which section relates to equivalence (or equality of area), of figures which are not identical

in form, not equal in all respects. The learner should not leave the first book before he has thoroughly understood and mastered the text: he will do well also to remember the different enunciations by the numbers of the propositions, which will not only assist him in calling to mind their order, which is frequently useful, but he will thereby be able to give the proper references to preceding propositions in the course of his demonstrations. It is too much to expect that he will at present be able to solve all the riders, but some he certainly should do, perhaps with a little assistance here and there.]