THEOREM XXXIX.

246. If two consecutive sides of one rectangle be respectively equal to two consecutive sides of another, the rectangles are con

HYPOTHESIS. Two quadrilaterals ABCD and FGHK, with all Zs rt.

(166. If interior angles on same side of transversal are supplemental, lines are

(210. If two polygons be mutually equilateral and mutually equiangular, they are

congruent.)

247. COROLLARY. A rectangle is completely determined by two consecutive sides; so if two sects, a and b, are given, we

may speak of the rectangle of a and b, or we may call it the rectangle ab. It can be constructed by the method given in 241 to describe a square.

THEOREM XL.

248. A parallelogram is equivalent to the rectangle of its base

AF

HYPOTHESIS. ABCD any □, of which side AD is taken as base;

altitude of .

CONCLUSION. ABCD rt. AFGD.

PROOF. AF = DG, and AB = DC.

(217. Opposite sides of a parallelogram are equal.)

By construction, ☀ G and ☀ F are rt.,

.. ▲ ABF ≈ ▲ DCG.

(179. Right triangles are congruent if hypothenuse and one side are respectively equal in each.)

From the trapezoid ABGD take away ▲ DCG, and then is left ABCD. From the same trapezoid take the equal ▲ ABF, and there is left the rt. — AFGD.

:: □ABCD = rt. — AFGD.

(89. If equals be taken from equals, the remainders are equal.)

249. COROLLARY. All parallelograms having equal bases and equal altitudes are equivalent, because they are all equivalent to the same rectangle.

EXERCISES. 50. Equivalent parallelograms on the same base and on the same side of it are between the same parallels.

EXERCISES. 51. Prove 248 for the case when C and F coin

cide.

52. If through the vertices of a triangle lines be drawn parallel to the opposite sides, and produced until they meet, the resulting figure will contain three equivalent parallelograms.

53. On the same base and between the same parallels as a given parallelogram, construct a rhombus equivalent to the parallelogram.

54. Divide a given parallelogram into two equivalent parallelograms.

55. Of two parallelograms between the same parallels, that is the greater which stands on the greater base. Prove also an inverse of this.

56. Equivalent parallelograms situated between the same parallels have equal bases.

57. Of parallelograms on equal bases, that is the greater which has the greater altitude.

58. A trapezoid is equivalent to a rectangle whose base is half the sum of the two parallel sides, and whose altitude is the perpendicular between them.

59. The sect joining the mid-points of the non-parallel sides of a trapezoid is half their sum.

60. If E and F are the mid-points of the opposite sides, AD, BC, of a parallelogram ABCD, the lines BE, DF, trisect the diagonal AC.

61. Any line drawn through the intersection of the diagonals of a parallelogram to meet the sides, bisects the surface.

62. The squares described on the two diagonals of a rhombus are together equivalent to the squares on the four sides. 63. Bisect a given parallelogram by a line passing through any given point.

64. In 244, what two rotations might be substituted for the two translations?

THEOREM XLI.

ALTERNATIVE PROOF.

250. All parallelograms having equal bases and equal altitudes are equivalent.

HYPOTHESIS. Two s with equal bases and equal altitudes. CONCLUSION. They can be cut into parts congruent in pairs. CONSTRUCTION. Place the parallelograms on opposite sides of their coincident equal bases, AB. Produce a side, as FB, which when continued will enter the other parallelogram. If it cuts out of the parallelogram at H before reaching the side CD opposite AB, then will the other cutting side, as CB, when produced, also leave the ABFG before reaching the side FG opposite AB; that is, the point, K, where CB cuts the line AG will be on the sect AG.

For, through H and K draw HL and KM || AB.

Then, in As ABH and ABK

XI=X1,

¥ 1 = ¥1, ¥ 2
× 1, X 2 = ¥ 2,

(168. If a transversal cuts two parallels, the alternate angles are equal.)

and side AB is common;

.. Δ ΑΒΗ Σ Δ ΑΒΚ.

..

But altitude of ▲ AHB < altitude of

ABCD,

altitude of ▲ ABK < altitude of □ ABFG, and K is on sect AG. Now, ▲ ABH≈ ▲ BHL, and ▲ ABK ≈ ▲ BKM.

Δ

(217. The diagonal of a parallelogram makes two congruent triangles.)

Taking away these four congruent triangles, we have left two parallelograms, HLCD and KMFG, with equal bases, and equal but diminished altitudes. Treat these in the same way as the parallelograms first given; and so continue until a produced side, as FRQ, and so the other, CRN, also, reaches the side opposite the base before leaving the parallelogram.

Then, as before, the As FRN and QRC are mutually equiangular ; but now we know their corresponding altitudes are equal.

and therefore the remaining trapezoids, PRQD ≈ PRNG.

(210. If two polygons be mutually equilateral and mutually equiangular, they are congruent.)

251. COROLLARY. Since a rectangle is a parallelogram, therefore a parallelogram is equivalent to the rectangle of its base and altitude.

EXERCISES. 65. How do you know that HLCD and KMFG have equal altitudes?

66. How do you know, that, if Q is on the sect CD, N is on the sect FG?