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Cor. 2. Also the lines drawn to the middle points of the opposite sides of a trapezium mutually bisect each other, since they are the diagonals of the parallelogram mentioned in Cor. 1.


THEOREM. Triangles, which have an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular.

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Let ABC, DFG be two triangles, having the angle A equal to the angle D, and the sides AB, AC proportional to the sides DF, DG; then will the triangle ABC be equiangular with the triangle DFG.

For, make DH equal to AB, and DK equal to AC, and join HK. Then the two triangles ABC, DHK, having two sides and the contained angle of the one equal to the two sides and the contained angle of the other, are equal (B. I, Prop. 11), and the angle B is equal to the angle DHK, the angle C to the angle DKH. The sides DH, DK, being respectively equal to AB, AC, are proportional to DF, DG; therefore HK is parallel to FG (B. IV, Prop. 11). Consequently the angles DHK, DKH are respectively equal to DFG, DGF (B. I, Prop. xx); but the angles DHK and DKH have just been shown to be respectively equal to the angles B and C, and consequently the angles DFG and DGF are respectively equal to the angles B and C.

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THEOREM. Two triangles having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles.

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Let the two triangles ABC, ADF have the angle A common; then will the

triangle ABC triangle ADF: AB.AC: AD. AF.

For, joining D and C, we have, since triangles of the same altitude are to each other as their bases,

triangle ABC triangle ADC AB AD, and

triangle ADC : triangle ADF :: AC : AF.

Multiplying together the corresponding terms of these proportions, and omitting the common term, triangle ADC, which enters into the antecedent and consequent of the first couplet, we have

triangle ABC : triangle ADF :: AB.AC: AD.AF.


THEOREM. Equiangular or similar triangles are to each other as the squares of their homologous sides.

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Let ABC, DFG be two equiangular triangles, AB and DF being two like sides; then will the triangle ABC be to the triangle DFG, as the square of AB is to the square of DF, or as AB2 to DF2.

For, the triangles being similar, they have their like sides proportional (B. IV, Prop. vII), and are to each other as the rectangles of the like pairs of their sides (B. IV, Prop. x1). Therefore,





therefore, taking the product of the corresponding terms of these proportions, we have

AB2: DF2: : AB.AC: DF.DG. But the

triangle ABC triangle DFG: AB.AC: DF.DG.

[B. IV, Prop. xI.]

Therefore triangle ABC : triangle DFG :: AB' : DF3.


THEOREM. A line which bisects any angle of a triangle, divides the opposite side into two segments, which are proportional to the two other adjacent sides.

Let the angle ACB of the triangle ABC be bisected by the line CD, making the angle ACD equal to the angle DCB; then will the segment AD be to the segment DB, as the side AC is to the side BC.



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For, let BF be drawn parallel to CD, meeting AC produced at the point F; then, because the line BC meets the two parallel lines CD, FB, it makes the angle CBF equal to the alternate angle BCD (B. I, Prop. xvii). Again, because AF meets the two parallel lines CD, FB, the exterior angle ACD is equal to the interior and opposite angle CFB (B. I, Prop. xx). Now since the angles BCD and ACD are equal, it follows that the angle CBF is equal to CFB; therefore the triangle CBF is isosceles, and the side CF is equal to CB (B. I, Prop. ví). Now, in the triangle ABF,-since CD is parallel to BF, we have

AD: DB:: AC: CF; [B. IV, Prop. 1.]

or, substituting CB for its equal CF,

AD: DB:: AC : CB.

(72.) We may also show the following proposition to be true:

THEOREM. If the exterior angle, formed by producing one of the sides of a triangle, be bisected by a line which meets the opposite side or base produced, then will the sides of the triangle be to each other" as the distances from the extremities of the base to the point where the bisecting line meets the base produced.

In the triangle ABC, let the side AC be produced, forming the exterior angle BCD : let CF bisect this angle, and meet the base produced at F; then will





For, through B draw BG parallel to CF; then will the angle CBG = = BCF (B. I, Prop. xvi), also CGB = DCF (B. I, Prop. xx); and since, by hypothesis, the angle BCF = - DCF, we must have... CBG CGB. Therefore the triangle CBG is isosceles, and CG=CB (B. I, Prop. vII).

Again, since BG is parallel to FC, we have AC : GC :: AF : BF (B. IV, Prop. 1); and substituting BC for GC, its equal, we have AC: BC: AF: BF.


THEOREM. If lines be drawn from the vertices of a triangle to bisect the opposite sides, they will mutually trisect each other:

In the triangle ABC, let CD, AF, BG be lines drawn from the vertices to bisect the opposite sides; then will they mu

tually trisect each other, that






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