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EXERCISES. 1. A B C is a triangle inscribed in a circle, and B D is drawn to meet the tangent to the circle at A in D, at an angle A BD equal to the angle A B C; show that A C is a fourth proportional to the lines B D, A D, A B.

2. Show that either of the sides of an isosceles triangle is a mean proportional between the base and the half of the segment of the base, produced if necessary, which is cut off by a straight line drawn from the vertex at right angles to the equal side.

3. A B is the diameter of a circle, D any point in the circumference, and C the middle point of the arc A D. If A C, A D, BC be joined and A D cut B C in E, show that the circle circumscribed about the triangle A E B will touch A C and its diameter will be a third proportional to B C and A B.

4. From the obtuse angle of a triangle draw a line to the base, which shall be a mean proportional between the segments into which it divides the base.

5. Find the point in the base produced of a right triangle, from which the line drawn to the angle opposite to the base shall have the same ratio to the base produced which the perpendicular has to the base itself.

6. A line touching two circles cuts another line joining their centres ; show that the segments of the latter will be to each other as the diameters of the circles.

7. Required the locus of the middle points of all the chords of a circle which pass through a fixed point.

8. O is a fixed point from which any straight line is drawn meeting a fixed straight line at P; in 0 P a point Q is taken such that 0 Q is to 0 P in a fixed ratio. Determine the locus of Q.

9. O is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P; in 0 P a point Q is taken such that 0 Q is to 0 P in a fixed ratio. Determine the locus of Q.

BOOK IV. COMPARISON AND MEASUREMENT OF THE SUR

FACES OF POLYGONS.

PROPOSITION I. THEOREM. 313. Two rectangles having equal altitudes are to cach other as their bases.

-C D

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Let the two rectangles be AC and A F, having the the same altitude AD.

rect. AC We are to prove

A B rect. A F A E

CASE I. — When A B and A E are commensurable.
Find a common divisor of the bases A B and A E, as A 0.

Suppose A O to be contained in A B seven times and in
A E four times.
Then

A B7

At the several points of division on A B and A E erect Is.

The rect. A C will be divided into seven rectangles,

and rect. A F will be divided into four rectangles.

These rectangles are all equal, for they may be applied to each other and will coincide throughout.

· rect AC 7
rect AF4

АВ

But

rect AC AB
rect A F = AE

CASE II. — When A B and A E are incommensurable.

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KE

Divide A B into any number of equal parts, and apply one of these parts to A E as often as it will be contained in A E.

Since A B and A E are incommensurable, a certain number of these parts will extend from A to a point K, leaving a remainder K E less than one of these parts.

Draw K H Il to E F.
Since A B and A K are commensurable,
rect. A H A K

Case 1 rect. AC A B Suppose the number of parts into which A B is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to E.

The limit of A K will be A E, and the limit of rect. A H will be rect. A F.

A K
.. the limit of 1 will be

U , A E

AB' and the limit of rect. A H

will be rect. A F

. rect. A "

rect. A C Now the variables A K and rect. A H

AB

rect. A are always equal however near they approach their limits; .. their limits are equal, namely,

rect. AF A E

: $ 199 rect. AC A B

Q. E. D. 314. COROLLARY. Two rectangles having equal bases are to each other as their altitudes. By considering the bases of these two rectangles A D and A D, the altitudes will be A B and A E. But we have just shown that these two rectangles are to each other as A B is to A E. Hence two rectangles, with the sime base, or equal bases, are to each other as their altitudes.

ANOTHER DEMONSTRATION. Let A C and A'C' be two rectangles of equal altitudes.

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Let b and b', S and S stand for the bases and areas of these rectangles respectively.

Prolong A D and A' D'.

Take A D, D E, E F .... m in number and all equal, and A'D', D' E', E' F', F'G' .. .. n in number and all equal.

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Now we can prove by superposition, that if A F be > A' G', rect. A P will be > rect. A' P'; and if equal, equal; and if less, less.

That is, if mb be > nb', m S is > nS'; and if equal, equal ; and if less, less. Hence,

6:6:::S', Euclid's Def., § 272

Q. E. D.

PROPOSITION II. THEOREM. 315. Two rectangles are to each other as the products of their bases by their altitudes.

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Let R and R' be two rectangles, having for their bases b and U, and for their altitudes a and a'.

Raxo
We are to prove Ri= axbo

Construct the rectangle S, with its base the same as that of R and its altitude the same as that of R'. Then

§ 314

(rectangles having the same base are to each other as their altitudes);

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(rectangles having the same altitude are to each other as their bases).
By multiplying these two equalities together

R a X 6
R = a' Xbi

Q. E. D. 316. DEF. The Area of a surface is the ratio of that surface to another surface assumed as the unit of measure.

317. DEF. The Unit of measure (except the acre) is a square a side of which is some linear unit; as a square inch, etc.

318. DEF. Equivalent figures are figures which have equal areas.

Rem. In comparing the areas of equivalent figures the symbol (=) is to be read “equal in area.”

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