But the base HC, and the triangle AHC, are any equimultiples whatever of the base BC, and the triangle ABC; and the base CL and the triangle ALC are any equimultiples whatever of the bafe CD and the triangle ADC; whence the base BC is to the base CD, as the triangle ABC is to the triangle ACD (V. Def. 5.)

Again, because the parallelogram CE is double the triangle ABC (I. 32.), and the parallelogram CF is double the triangle ADC, the triangle ABC will be to the triangle ADC as the parallelogram CE is to the parallelogram CF (V. 12.)

But, it has been fhewn, that the bafe BC is to the base CD, as the triangle ABC is to the triangle ADC; therefore the base BC is alfo to the bafe CD, as the parallelogram CE is to the parallelogram cr. Q. E. D.

COROLL. Triangles and parallelograms, having equal altitudes, are to each other as their bafes.

PRO P. II.

Triangles and parallelograms, having equal bafes, are to each other as their altitudes.

A

Let ABC, DEF be two triangles, having the equal bases AB, DE, and whofe altitudes are CH, FI; then will the

triangle

triangle ABC have the same ratio to the triangle DEF, as CH has to FI.

For make BP perpendicular to AB, and equal to CH (I. 11 and 3.); and in BP take BQ equal to FI, and join AP, AQ and CP.

Then, because BP is equal to CH, and the base AB is common, the triangle ABP will be equal to the triangle ABC (II. 5.)

And, because AB is equal to DE, and BQ to FI, the triangle ABQ will also be equal to the triangle DEF (II. 5.)

But the triangle ABP is to the triangle ABQ as BP is to BQ (VI. 1.); therefore the triangle ABC is also to the triangle DEF as BP is to BQ, or as CH to FI (V.

9.)

And, fince parallelograms, having the fame bafes and altitudes, are the double of these triangles, they will, likewife, have to each other the same ratio as their altitudes.

Q. E. D.

COR. 1. If the bases of equal triangles are equal, the altitudes will also be equal; and if the altitudes are equal, the bafes will be equal.

COR. 2. From this, and the former propofition, it also appears, that rectangles, which have one fide in each equal, are proportional to their other fides.

PROP. fit. THEOREM.

If a right line be drawn parallel to one of the fides of a triangle, it will cut the other fides proportionally and if the fides be cut proportionally, the line will be parallel to the remaining fide of the triangle.

Let ABC be a triangle, and DE be drawn parallel to the fide BC; then will AD be to DB, as AE is to EC. For join the points B, E, and C, D:

Then, because the triangles DBE, DCE are upon the fame base DE, and between the fame parallels DE, BC, they will be equal to each other (I. 31.)

And, fince equal magnitudes have the fame ratio to the fame magnitude (V. 9.), the triangle DBE will be to the triangle DAE, as the triangle DCE is to the triangle DAE.

But triangles of the same altitude are to each other as their bafes (VI. 1.); whence the triangle DBE will be to the triangle DAE as DB is to`DA.

For the fame reason, the triangle DCE will be to the triangle DAE, as EC is to EA.

And, fince ratios which are the fame to the fame ratio, are the fame to each other (V. 11.) DB will be to DA as EC is to EA; or, inversely, AD to DB as AE to EC.

Again, let the fides AB, AC be cut proportionally, in the points D and E; then will the line DE be parallel

to BC.

For, the fame construction being made as before, the triangle DEB will be to the triangle DEA, as DB is to DA (VI. 1.); and the triangle EDC to the triangle DEA as EC to EA (VI. 1.)

And, fince DB is to DA as EC is to EA (by Conft.), the triangle DEB will be to the triangle DEA as the triangle EDC is to the triangle DEA (V. 11.)

But magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.); whence the triangle DEB is equal to the triangle EDC.

And fince these triangles are equal to each other, and are upon the fame base DE, they will have equal altitudes (VI. 2. Cor.), or ftand between the fame parallels ; whence DE is parallel to BC.

Q.E.D.

COROLL. In the fame manner it may be fhewn, that the fides of the triangle are proportional to any two of the parts into which they are divided; and that the like parts of each are alfo proportional.

If the verticle angle of a triangle be bifected, the fegments of the bafe will have the fame ratio with the other two fides: and if the fegments have the fame ratio with the other two fides the angle will be bifected.

Let the angle BAC of the triangle ABC be bifected by the right line AD; then will BD be to DC as BA is to AC.

For through the point c draw CE parallel to DA (I. 27.), and let BA be produced to meet CE in E:

Then, because the right line AC cuts the two parallel right lines AD, EC, the angle ACE will be equal to the alternate angle CAD (I. 24.)

But the angle CAD is equal to the angle BAD, by the propofition; therefore the angle BAD is alfo equal to the angle ACE.

And, in like manner, because the right line BE cuts the two parallel right lines AD, EC, the outward angle BAD will be equal to the inward opposite angle AEC (I. 25.) M 2

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