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has to the second. And consequently, if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line is given; and if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second is given.

Let AC, DF, be two equiangular parallelograms; as BC, a side of the first, is to EF, a side of the second, so is DE, the other side of the second, to the straight line to which AB, the other side of the first, has the same ratio which AC has to DF.

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Produce the straight line AB, and make as BC to EF so DE to BG, and complete the parallelogram BGHC; therefore, because BC or GH, is to EF, as DE to BG, the sides about the equal angles BGH, DEF, are reciprocally proportional; wherefore*, the parallelogram BH is equal to DF; and AB is to BG, as the parallelogram AC is to BH, that is, to DF; as therefore BC is to EF, so is DE to BG, which is the straight line to which AB has the same ratio that AC has to DF.

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And if the ratio of the parallelogram AC to DF be given, then the ratio of the straight line AB to BG is given; and if the ratio of AB to the straight line BG be given, the ratio of the parallelogram AC to DF is given.

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If two parallelograms have unequal, but given angles, and See N. if as a side of the first to a side of the second, so the other side of the second be made to a certain straight line; if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line shall be given. And if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second shall be given.

Let ABCD, EFGH, be two parallelograms which

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* 35. 1.

have the unequal but given angles ABC, EFG; and as BC to FG, so make EF to the straight line M. If the ratio of the parallelogram AC to EG be given, the ratio of AB to M is given.

At the point B of the straight line BC make the angle CBK equal to the angle EFG, and complete the parallelogram KBCL. And because the ratio of AC to EG is given, and that AC is equal to the parallelogram KC, therefore the ratio of KC to EG is given; and KC, EG, are equiangular, therefore as BC to FG, so is* EF to the straight line to which KB has a given ratio, viz. the same which the parallelogram KC has to EG; but as BC to FG, so is EF to the straight line M; therefore KB has a given ratio to M; and the ratio of AB to BK is given, because the triangle ABK * 43 Dat. is given in species*; therefore the ratio of AB to M is given*.

* 63 Dat.

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And if the ratio of AB to M be given, the ratio of the parallelogram AC to EG is given; for since the ratio of KB to BA is given, as also the ratio of AB to M, the ratio of KB to M is given*; and because the parallelograms KC, EG, are equiangular, as BC to * 63 Dat. FG, so is* EF to the straight line to which KB has the same ratio which the parallelogram KC has to EG: but as BC to FG, so is EF to M; therefore

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KB is to M as the parallelogram KC is to EG and the ratio of KB to M is given, therefore the ratio of the parallelogram KC, that is, of AC, to EG is given.

COR. And if two triangles ABC, EFG, have two equal angles, or two unequal but given angles ABC, EFG, and if as BC a side of the first to FG a side of the second, so the other side of the second EF be made to a straight line M; if the ratio of the triangles be given, the ratio of the other side of the first to the straight line M is given.

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Complete the parallelograms ABCD, EFGH; and because the ratio of the triangle ABC to the triangle EFG is given, the ratio of the parallelogram AC to EG is given*, because the parallelograms are double of the triangles; and because BC is to FG, as EF to M,

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the ratio of AB to M is given by the 63 Dat. if the angles ABC, EFG, are equal; but if they be unequal but given angles, the ratio of AB to M is given by this proposition.

And if the ratio of AB to M be given, the ratio of the parallelogram AC to EG is given by the same propositions; and therefore the ratio of the triangle ABC to EFG is given.

PROP. LXV.

If two equiangular parallelograms have a given ratio to one another, and if one side have to one side a given ratio; the other side shall also have to the other side a given ratio.

Let the two equiangular parallelograms AB, CD, have a given ratio to one another, and let the side EB have a given ratio to the side FD; the other side AE has also a given ratio to the other side CF.

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Because the two equiangular parallelograms AB, CD, have a given ratio to one another; as EB, a side of the first, is to FD, a side of the second, so is FC, 63 Dat. the other side of the second, to the straight line to which AE, the other side of the first, has the same given ratio which the first parallelogram AB has to the other CD. Let this

straight line be EG; therefore the ratio of AE to EG is given; and EB is to FD, as FC to EG, therefore the ratio of FC to EG is given, because the ratio of EB

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to FD is given; and

because the ratio of AE to EG, as also the ratio of FC to EG is given; the ratio of AE to CF is given*.

The ratio of AE to CF may be found thus: take a straight line H given in magnitude: and because the ratio of the parallelogram AB to CD is given, make the ratio of H to K the same with it. And because the ratio of FD to EB is given, make the ratio of K to L the same: the ratio of AE to CF is the same with the ratio of H to L. Make as EB to FD, so FC to EGatherefore, by inversion, as FD to EB, so is EG to FC; and as AE to EG, so is* (the parallelogram

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AB to CD, and so is) H to K: but as EG to FC, so is (FD to EB, and so is) K to L; therefore, ex æquali, as AE to FC, so is H to L.

PROP. LXVI.

If two parallelograms have unequal but given angles, and a given ratio to one another; if one side have to one side a given ratio, the other side has also a given ratio to the other side.

Let the two parallelograms ABCD, EFGH, which have the given unequal angles ABC, EFG, have a given ratio to one another, and let the ratio of BC to FG be given; the ratio also of AB to EF is given.

At the point B of the straight line BC make the angle CBK equal to the given angle EFG, and complete the parallelogram BKLC: and because each of the an43 Dat. gles BAK, AKB, is given, the triangle ABK is given

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in species; therefore the ratio of AB to BK is given;
and because, by the hypothesis, the ratio of the parallel-
ogram AC to EG is given, and that AC is equal to
BL; therefore the ratio of BL to EG is given: and
because BL is equiangular to EG, and, by the hypo-
thesis, the ratio of BC to
FG is given; therefore the
ratio of KB to EF is given,
and the ratio of KB to BA
is given; the ratio therefore*
of AB to EF is given.

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The ratio of AB to EF may be found thus: take the straight line MN given in position and magnitude; and

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make the angle NMO equal to the given angle BAK, and the angle MNO equal to the given angle EFG or AKB: and because the parallelogram BL is equiangular to EG, and has a given ratio to it, and that the ratio of BC to FG is given'; find by the 65th Dat. the ratio of KB to EF; and make the ratio of NO to OP the same with it: then the ratio of AB to EF is the same with the ratio of MO to OP: for since the triangle ABK is equiangular to MON, as AB to BK, so is MO to ON: and as KB to EF, so is NO to OP; therefore, ex æquali, as AB to EF, so is MO to OP.

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If the sides of two equiangular parallelograms have given See N. ratios to one another; the parallelograms shall have a given ratio to one another.

Let ABCD, EFGH, be two equiangular parallelograms, and let the ratio of AB to EF, as also the ratio of BC to FG, be given; the ratio of the parallelogram AC to EG is given.

Take a straight line K given in magnitude, and because the ratio of AB to EF is given, make the ratio of K to L the same with it; therefore L is given and because the ratio of BC to FG is given, make the ratio of L to M the same: therefore M is

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* 2 Dat.

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given, and K is given: wherefore the ratio of K to 2 Dat. M is given: but the parallelogram AC is to the parallel- * 1 Dat. ogram EG, as the straight line K to the straight line M, as is demonstrated in the 23d Prop. of B. 6. Elem.; therefore the ratio of AC to EG is given.

From this it is plain how the ratio of two equiangular parallelograms may be found when the ratios of their sides are given.

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If the sides of two parallelograms which have unequal, but See N. given angles, have given ratios to one another; the parallelograms shall have a given ratio to one another.

Let two parallelograms ABCD, EFGH, which have the given unequal angles ABC, EFG, have the ratios of their sides, viz. of AB to EF, and of BC to FG, given; the ratio of the parallelogram AC to EG is given.

At the point B of the straight line BC, make the angle CBK equal to the given angle EFG, and complete the parallelogram KBCL: and because each of the angles BAK, BKA, is given, the triangle ABK is given in species: therefore the ratio of AB to BK is * 43 Dat.

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