236. The intersection point of medials is called the Centroid of the triangle. The intersection point of perpendiculars from the vertices to the opposite sides is called the Orthocenter of the triangle. The intersection point of perpendiculars erected at the mid-points of the sides is called the Circumcenter of the triangle.

IV. Equivalence.

237. Two plane figures are called equivalent if we can prove that they must contain the same extent of surface, even if we do not show how to cut them into parts congruent in pairs.

238. The base of a figure is that one of its sides on which we imagine it to rest.

Any side of a figure may be taken as its base.

239. The altitude of a figure is the perpendicular from its highest point to the line of its base.

So the altitude of a parallelogram is the perpendicular dropped from any point of one side to the line of the opposite side.

240. The words "altitude" and "base" are correlative; thus, a triangle may have three different altitudes.

PROBLEM XI

241. To describe a square upon a given sect.

GIVEN, the sect AB.

REQUIRED, to describe a square on AB.

CONSTRUCTION. By 135, from A draw ACL AB.
By 133, in AC make AD = AB.

By 167, through D draw DF || AB.

AF will be the required square.

PROOF. By construction, AF is a parallelogram,

.*.

By 167, through B draw BF || AD.

AD = BF,

X D.

AB = FD, and

X F = XA, and B

(217. In a parallelogram the opposite sides and angles are equal.)

But, by construction, AB = AD,

.. AF is equilateral.

Again, A + ¥ B = st. ¥.

(169. If two parallel lines are cut by a transversal, the two interior angles are supplemental.)

242. COROLLARY. If a parallelogram have one angle right,

it is a rectangle.

THEOREM XXXVII.

243. In any right-angled triangle, the square on the hypothenuse is equivalent to the sum of the squares of the other two sides.

HYPOTHESIS. ▲ ABC, right-angled at B.

CONCLUSION. Square on AB + square on BC = square on AC. PROOF. By 241, on hypothenuse AC, on the side toward the ▲ ABC, describe the square ADFC.

On the greater of the other two sides, as BC, by 133, lay off CG = AB. Join FG.

Then, by construction, CA = FC, and AB = CG, and X CAB = FCG, since each is the complement of ACB;

.. ▲ ABC ≈ ▲ CGF.

Translate the ▲ ABC upward, keeping point A on sect AD, and point C on sect CF, until A is on D, and C on F; then call B' the position of B.

Likewise translate CGF to the right until C is on A, and F on D; then call G' the position of G.

The resulting figure, AG'DB'FGBA, will be the squares on the other two sides, AB and BC.

For, since the sum of the angles at D = st. 4,

:. G'D and DB' are in one line.

244. COROLLARY. Given any two squares placed side by side, with bases AB and BC in line; to cut this figure into three pieces, two of which being translated without rotation, the figure shall be transformed into one square.

In AC take AD = BC, and join D to the corners of the squares opposite B. Two right-angled triangles are thus produced, with hypothenuses perpendicular to one another. Translate each triangle along the hypothenuse of the other.

THEOREM XXXVIII

245. Two triangles are congruent if they are mutually equi

angular, and have corresponding altitudes equal.

HYPOTHESIS. AS ABC and BDF mutually equiangular, and altitude BH = altitude GB.

CONCLUSION. ▲ ABC ≈ ▲ BDF.
Δ Δ

PROOF. Rt. A BDG ≈ rt. ▲ BCH,

(176. Triangles are congruent when two angles and a side in one are equal to two angles and a corresponding side in the other.)

(124. Triangles are congruent when two angles and the included side are equal in each.)

EXERCISES. 47. If from the vertex of any triangle a perpendicular be drawn to the base, the difference of the squares on the two sides of the triangle is equal to the difference of the squares on the parts of the base.

48. Show how to find a square triple a given square.

49. Five times the square on the hypothenuse of a rightangled triangle is equivalent to four times the sum of the squares on the medials to the other two sides.