draw a straight line in the direction BC and produce it until it intersects at E the perpendicular erected upon CD. Now AC=CD, angle A = D (being R. A.,) and angle ACB = DCE, (107 ;) therefore triangle ACB= DCE, (127;) consequently AB=DE. The length of AB is found by measuring DE.

211. To divide a given triangle ABC, (fig. 112,) into two equivalent parts by a straight line.

Sol. 1. Draw the line AD from the vertex of one of the angles A to the middle of the opposite side. Now BD=DC; that is, the triangles ABD and ADC have equal bases, and since their altitudes are also equal, the triangles are equivalent, (153.)

Sol. 2. Suppose it is required to draw the dividing line from the point D in one of the sides AB of the given triangle. Bisect the side AB, (fig. 113,) at E. Draw CE and CD. Now the triangle BDC is greater than ABC by the part EDC; therefore we must take from triangle BDC a part equivalent to EDC. Draw EG parallel to DC; and then draw DG, which will be the division line required. For triangle EDC=CDG; since they have the same base CD, and an equal altitude. If therefore we take the triangle CDG from the triangle CDB, we shall take a part equivalent to EDC. There will remain the triangle BDG = BEC; and consequently = ABC. Thus the line DG divides the triangle ABC into two equivalent parts.

212. To transform a parallelogram, ABCD, (fig. 114,) into a triangle of equal area.

Sol. Produce one side of the parallelogram BC, taking CE=BC. Draw AE. The triangle ABE is the one required.

Dem. Draw AC. Triangle ABC=ACE and triangle ABC = ACD, (153;) therefore triangle ACE= ACD. Add triangle ABC to each side of this equation, and we have ACE + ABC (=ABE) = ACD + ABC (= parallelogram ABCD.) That is, if a triangle has an equal altitude and twice as great a base as a parallelogram, the triangle will be equivalent to the parallelo

gram.

213. To construct a triangle which shall be 11⁄2 times as great as a given parallelogram ABCD.

Sol. (Fig. 115.) Produce BC and take BE=3 BC. Draw BD and DE. BDE is the required triangle.

Dem. Draw DF. Triangle BCD = DCF = DFE, (153.) Triangle BCD = + parallelogram ABCD; therefore BDE=3 × 4 ABCD=1+ ABCD.

214. To transform a triangle into a parallelogram of equal area.

Sol. Construct a parallelogram upon the base of the triangle, with an altitude equal to that of the triangle. It can readily be demonstrated that such a parallelogram will be equivalent to the triangle.

If it is required that the parallelogram should have an angle of a given size; then bisect BC, (fig. 116,) at D. Make at D, in line DC, an angle equal to the given angle. Through A draw AF parallel to BC; and through C draw CF parallel to DE. DCEF is the parallelogram required.

215. In a parallelogram two intersecting diagonals bisect each other.

Dem. (Fig. 117.) AB= DC, (147;) angle EAB = ECD and angle EBA = EDC, (110,) consequently triangle AEB = DEC; therefore EB

= EC.

ED and EA

216. Two diagonals drawn in a square or a lozenge cut each other at right angles; in all other parallelograms at oblique angles.

Dem. Let ABCD, (fig. 118,) be a square or a lozenge; then AE=EC, (215,) DE=DE; AD=DC by supposition; consequently triangle AED=CED, (137;) therefore angle AED=CED; and consequently DE is perpendicular to AC (2.)

2. Let ABCD, (fig. 117,) be a rectangle or a rhomboid, in which the sides AB and DC are greater than AD and BC. AE=AE, ED = EB, (215,) but AB >AD; therefore angle AEB>AED, (144.) But angle AEB + AED=2 R. A. (106;) therefore AEB >1 R. A., and AED<1 R. A., that is, AC cuts DB obliquely.

217. Two diagonals drawn in a square or a lozenge divide the angles into equal parts ; in all other parallelograms into unequal parts.

Dem. Let ABCD, (fig. 118,) be a square or a lozenge; then AB=AD; AC=AC; BC = DC; therefore triangle ABC=ADC; consequently angle CAB=CAD.

In parallelogram ABCD, (fig. 117,) side AB>BC; therefore angle ACB>CAB, (138.) But angle ACB = CAD, (110 ;) therefore angle CAD>CAВ.

218. To divide a parallelogram into 2, 3, 4, &c. equal parts.

Sol. (Fig. 119.) Divide any side AB into the required number of parts, and through the division points draw lines parallel to the side AD; the given parallelogram will thus be divided as was required.

219. Let GF (fig. 120) be a straight line drawn through the middle of a diagonal of a parallelogram; how will it divide the parallelogram?

Sol. Triangle ADC = ABC, (146.) EC = EA, angle ECG = EAF and CEG = AEF, (107,) therefore triangle EGC = EFA. Consequently triangle ADC - EGC (= quadrilateral AEGD)=triangle ABC

EFA (= quadrilateral ECBF.) Again, triangle EGC = EFA, therefore AEGD + EFA = AFGD= ECBF + EGC = BFGC; that is, the parallelogram ABCD is divided into two equal parts.

220. Let ABC (fig. 121) be a right triangle of which B is the right angle. Upon each side of the triangle construct a square. What ratio will the square constructed upon the hypothenuse bear to the squares constructed upon the sides which include the right angle?

Sol. If the side AB = BC, the squares constructed upon those sides will be equal, and a perpendicular let fall from the vertex of the angle B upon AC will bisect AC, (133,) and if produced will bisect the square ACDE. If AB>BC, then square ABFG>BCKH, and the perpendicular let fall from the vertex of the angle B will divide AC and also the square ACDE into 2 unequal parts; the greater part being nearest to A.

Let fall upon AC the perpendicular BL, and produce it to M. Draw BD and AK. The triangle ACK= ВСКН, (152.) Triangle BCD = + CLMD. In triangles ACK and BCD we have AC=CD and CK= CB; angle BCK=ACD being right angles and BCA =BCA; therefore angle BCK + BCA (= angle ACK) = ACD + BCA (= BCD;) therefore triangle ACK = BCD, (126,) consequently twice the triangle ACK (= BCKH) is equal to twice the triangle BCD (= CLMD.) It may be demonstrated in a similar manner, that the parallelogram ALME = square ABFG. Consequently CDML + LMEA = square ACED = square AFGB + BCKH. That is, in every right-angled triangle the square of the hypothenuse is equal to the sum of the squares of the other two sides. This is called the proposition of Pythagoras, because he first discovered the ratio.

221. Cor. 1. The square of the hypothenuse of a right-angled isosceles triangle is equivalent to twice the square of one of the equal sides.

Cor. 2. A square circumscribed about a circle is double a square inscribed in the same circle. For the side of the former is equal to a diagonal of the latter, and is consequently the hypothenuse of a right-angled isosceles triangle.

222. To construct a square which shall be equal to 2 given squares.

Sol. (Fig. 122.) Let M and N be the sides of the given squares. Construct a right angle; take one of its sides equal to M, and the other side equal to N. Connect the extremities of these sides by the line X. The square constructed upon X will equal the sum of the squares of the sides M and N.

223. If it is required to construct a square which shall be equal to the sum of 3, 4, 5 or more squares, a similar method is adopted. Suppose, for example, the sides of the given squares are M, N, O, P, and Q, (fig.