PROPOSITION XIV. THEOREM.

Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles -reciprocally proportional.

Let AB, BC be equals, having ▲ FBD= ▲ EBG.

Then must DB be to BE as GB is to BF.

Place thes so that DB and BE are in the same st. line; then must GB and BF also be in one st. line.

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Parallelograms, which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let the sides about the equal 4s be reciprocally proportional, that is, let DB be to BE as GB is to BF.

Then must ☐ AB=□BC.

For, the same construction being made,

DB is to BE as GB is to BF,

PROPOSITION XV. THEOREM.

Equal triangles, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional.

Let ABC, ADE be equal As, having 4 BAC=1 DAE.

Then must CA be to AD as EA is to AB.

Place the As so that CA and AD are in the same st. line; then must EA and AB also be in one st. line.

Join BD.

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Then . ▲ ABC= ▲ ADE, and ABD is another ▲,

.. AABC is to A ABD as A ADE is to A ABD.

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VI. 1.

VI. 1.

V. 5.

But as ▲ ABC is to ▲ ABD so is CA to AD, and as AADE is to A ABD so is EA to AB.

.. CA is to AD as EA is to AB.

Ex. 1. Shew that, provided the sides of one of the triangles be made the extremes, it is indifferent, so far as the truth of the Proposition is concerned, in what order the sides of the other triangle are taken as the means of the four proportionals.

Ex. 2. ABb, AcC are two given straight lines, cut by two others BC, bc, so that the two triangles ABC, Abc may be equal; shew that the lines BC, bc divide each other proportionally.

And Conversely,

Triangles, which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let the sides about the equals be reciprocally proportional, that is, let CA be to AD as EA is to AB.

Then must ▲ ABC= ^ ADE.

For, the same construction being made,

CA is to AD as EA is to AB,

and that CA is to AD as ▲ ABC is to ▲ ABD,
and that EA is to AB as A ADE is to ▲ ABD,.
..A ABC is to ▲ ABD as ▲ ADE is to ▲ ABD.
and ... A ABC= ▲ ADE.

VI. 1.

VI. 1.

V. 5.

V. 8.

Q. E. D.

Ex. 3. Through the extremities of the base BC, of a triangle ABC, draw two parallel lines, BE and CD, meeting AC and AB produced in E and D respectively, so that BCD may be equal in area to ABE.

Ex. 4. P is any point on the side AC, of the triangle ABC; CQ, drawn parallel to BP, meets AB produced in Q; AN, AM are mean proportionals between AB, AQ, and AC, AP, respectively. Shew that the triangle ANM is equal to the triangle ABC.

PROPOSITION XVI. THEOREM.

If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means..

Let the four st. lines AB, CD, EF, GH be proportionals, so that AB is to CD as EF is to GH.

Then must rect. AB, GH=rect. CD, EF.

Draw AML to AB, and CN1 to CD; and make AM=GH, and CN=EF;

and complete the s BM, DN. Then ·.· AB is to CD as EF is to GH, and that EF=CN, and GH=AM,

.. AB is to CD as CN is to AM.

Thus the sides about the equals of the Os BM, DN are reciprocally proportional,

and.. BM=□ DN ;

that is, rect. AB, AM=rect. CD, CN.

..rect. AB, GH =rect. CD, EF.

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V. 6.

equiangular

VI. 14.

Ex. 1. If E be the middle point of a semicircular arc AEB, and EDC be any chord, cutting the diameter in D, and the circle in C, prove that the square on CE is equal to twice the quadrilateral AEBC.

And Conversely,

If the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals.

Let rect. AB, GH=rect. CD, EF.

Then must AB be to CD as EF is to GH.

For, the same construction being made,

rect. AB, GH =rect. CD, EF,

.. rect. AB, AM=rect. CD, CN,

that is, BM=□ DN.

and theses are equiangular to one another,

and the sides about the equals are reciprocally

proportional,

and.. AB is to CD as CN is to AM,

and.. AB is to CD as EF is to GH.

VI. 14.

V. 6.

Q. E. D.

Ex. 2. If, from an angle of a triangle, two straight lines be drawn, one to the side subtending that angle, and the other cutting from the circumscribing circle a segment, capable of containing an angle, equal to the angle, contained by the first drawn line and the side, which it meets; the rectangle, contained by the sides of the triangle, shall be equal to the rectangle, contained by the lines thus drawn.