And if, to these equals, there be added the angle ABC, the angles CBD, ABC, taken together, will be equal to the three angles BCA, CAB and ABC, taken together. But the angles CBD, ABC, taken together, are equal to two right angles (Prop. 13.); consequently the three angles BCA, CAB and ABC, taken together, are also equal to two right angles. COROLL. I. If two angles of one triangle, be equal to two angles of another, each to each, the remaining angles will also be equal. COROLL. 2. Any quadrilateral may be divided into two triangles; therefore all the four angles of such a figure, taken together, are equal to four right angles. PRO P. XXIX. THEOREM. Right lines joining the corresponding extremes of two equal and parallel right lines are themselves equal and parallel. Let AB, DC be two equal and parallel right lines; then will the right lines AD, BC, which join the corresponding extremes of those lines, be also equal and parallel. For draw the diagonal, or right line AC : Then, because AB is parallel to Dc, and ac intersects them, the angle DCA will be equal to the alternate angle CAB (Prop. 24.) And D 3 And, because AB is equal to DC (by Hyp.), Ac common to each of the triangles ABC, ADC, and the angle DCA equal to the angle CAB, the fide AD will also be equal to the side BC, and the angle dac to the angle ACB (Prop. 4.) Since, therefore, the right line ac intersects the two right lines AD, BC, and makes the alternate angles equal to each other, those lines will be parallel (Prop.23.) But the line Ad has been proved to be equal to the line BC; consequently they are both equal and parallel. Q.E.D. PRO P. XXX. THEOREM. The opposite sides and angles of any parallelogram are equal to each other, and the diagonal divides it into two equal parts. D Let ABCD be a parallelogram, whose diagonal is AC; then will its opposite sides and angles be equal to each other, and the diagonal AC will divide it into two equal parts. For, since the fide ad is parallel to the side bc (Def. 22.), and the right line ac intersects them, the angle DAC will be equal to the alternate angle ACB (Prop. 24.) And, because the side pc is parallel to the side AB (Def. 22.), and ac intersects them, the angle Dca will be equal to the alternate angle CAB (Prop. 24.) Since, therefore, the two angles DAC, DCA, are equal to the two angles ACB, CAB, each to each, the remain ing angle ADC will also be equal to the remaining angle ABC (Prop. 29. Cor.) and the whole angle DAB to the whole angle DCB. But, the triangles CDA, ABC, being mutually equiangular, and having ac common, the side Dc will also be equal to the side AB, and the fide ad to the fide bc, and the two triangles will be equal in all respects (Prop. 21.) R.E. D. PRO P. XXXI. THEOREM. Parallelograms, and triangles, standing upon the same base, and between the fame parallels, are equal to each other, Let AE, Bd be two parallelograms ftanding upon the same bafe AB, and between the fame parallels AB, DE; then will the parallelogram AE be equal to the parallelogram BD. For, since ad is parallel to BC (Def. 22.), and be intersects them, the outward angle ECB will be equal to the inward opposite angle FDA (Prop. 25.) And, because AF is parallel to Be (Def. 22.), and de intersects them, the outward angle AFD will be equal to the inward opposite angle bec (Prop. 25.) Since, therefore, the angle ecb is equal to the angle FDA, and the angle AFD to the angle Bec, the remaining D4 angle angle cbe will be equal to the remaining angle dar (Prop. 29. Cor. 1.) But the side Ad is also equal to the side BC (Prop. 31.); consequently, since the triangles ADF, Bce are mutually equiangular, and have two corresponding fides equal to each other, they will be equal in all respects (Prop. 21.) If, therefore, from the whole figure ABED, there be taken the triangle BCE, there will remain the parallelogram BD; and if, from the same figure, there be taken the triangle ADF, there will remain the parallelogram AE. But if equal things be taken from the same thing, the remainders will be equal ; consequently, the parallelogram AE is equal to the parallelogram BD.. Again, let ABC, ABF be two triangles, standing upon the same base AB, and between the same parallels, AB, CF; then will the triangle ABC be equal to the triangle ABF. For produce cf, both ways, to D and E, and draw AD parallel to Bc, and be to AF (Prop. 28.) Then, since BD, AE, are two parallelograms, standing upon the same base AB, and between the fame parallels AB, DE, they are equal to each other (Prop. 32.) And, because the diagonals AC, BF bisect them (Prop. 31.), the triangle ABC will also be equal to the triangle ABF. Q. E. D. PRO P. XXXII. THEOREM. If a parallelogram and a triangle stand upon the saine base, and between the same parallels, the parallelogram will be double the triangle. E Let the parallelogram ac and the triangle AEB stand upon the same base AB, and between the same parallels AB, DE; then will the parallelogram Ac be double the triangle Aeb. For join the points B, D; then will the parallelogram AC be double the triangle ADB, because the diagonal DB divides it into two equal parts (Prop. 31.) But the triangle ADB is equal to the triangle AEB, because they stand upon the same base AB, and between the same parallels AB, de (Prop. 32.); whence the parallelogram ac'is also double the triangle AEB. E. D. COROLL. If the base of the parallelogram be half that of the triangle, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other. PRO P. |