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Again, if CD be taken so that A and D are on the same side of C, the like construction will determine the external point of division.
In this case the construction will fail if CD = AC, for D would coincide with A.
As above, we may also prove that there can be only one point of external division in the given ratio.
503. INVERSE OF 499.
A line which divides two sides of a triangle proportionally is parallel to the third.
For a parallel from one of the points would divide the second side in the same ratio, but there is only one point of division of a given sect in a given ratio.
504. Rectangles of equal altitude are to one another in the same ratio as their bases.
Let AC, BC, be two rectangles having the common side OC, and their bases OA, OB, on the same side of OC.
In the line OAB take OM = m.OA, and ON = n.OB, and complete the rectangles MC and NC.
Then MC = m.AC, and NC = n.BC; and as OM is >, =, or < ON, so is MC respectively >, =, or < NC;
:: rectangle AC : rectangle BC :: base OA : base OB.
505. COROLLARY. Parallelograms or triangles of equal altitude are to one another as their bases.
506. In the same circle, or in equal circles, angles at the center and sectors are to one another as the arcs on which they stand.
Let O and C be the centers of two equal circles ; AB, KL, any two arcs in them.
Take an arc AM = m. AB; then the angle or the sector between OA and OM equals m .AOB.
(365. In equal circles, equal arcs subtend equal angles at the center.)
Also take an arc KN = n.KL; then the angle or sector between CK and CN equals n. KCL.
But as AOM>, =, or < KCN, so respectively is arc AM>,=, or < arc KN; (370 and 372. In equal circles, equal angles at the center or equal sectors intercept equal arcs, and of two unequal angles or sectors the greater has the greater arc.)
... AOB : KCL :: arc AB : arc KL.
II. Similar Figures.
507. Similar figures are those of which the angles taken in the same order are equal, and the sides between the equal angles proportional.
The figure ABCD is similar to A'B'C'D if X A = * A', % B= * B', etc., and also
AB : A'B :: BC: B'C' :: CD : C'D :: etc.
AS ABC, FGH, having s at A, B, and C, XS at F, G, and H.
CONCLUSION. AB : FG :: BC : GH :: CA : HF.
PROOF. Apply A FGH to A ABC so that the point G coincides with B, and GH falls on BC. Then, because x G = * B,
GF falls on BA.
(166. If corresponding angles are equal, the lines are parallel.)
therefore, by 498, AB : BF :: CB : BH'.
In the same way, by applying the A FGH so that the Xs at H and C coincide, we may prove that BC : GH :: CA : HF.
A triangle is similar to any triangle cut off by a line parallel to one of its sides,
510. Triangles having their sides taken in order proportional are similar.
HYPOTHESIS. AB : FG :: BC: GH :: CA : HF.
.. AB : FB :: BC: BH' :: CA : H'F'.
(508. Mutually equiangular triangles are similar.)
Since FG = FB, :. BC : GH :: BC : BH',
(484. Ratios equal to the same ratio are equal.)
therefore, by 491,
BH'. In the same way
HF = H'F',
A FGH SA FBH'. But FBH' is similar to ABC.
511. Two triangles having one angle of the one equal to one angle of the other, and the sides about these angles proportional, are similar.
HYPOTHESIS. B = 4G, and AB : BC :: FG : GH. CONCLUSION. A ABC ~AFGH (using ~ for the word “similar '). PROOF. In BA take BF = GF, and draw F'L || AC,
.:. Δ FBL ΝΔ ABC,
(508. Equiangular triangles are similar.)
:. AB : BC :: FB: BL.
But FB = FG by construction; therefore, from our hypothesis,
(124. Triangles having two sides and the included angle respectively equal are