Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

SCHOLIUM.

WE will here (a) shew by way of Anticipation, the truth of the Pythagorick Theorem, esteemed worth an He∣catomb: Which hereafter we will demonstrate after other dif∣ferent ways; viz. In a Right Angled Triangle BAC (Fig. 29.) the Square of the greatest Side opposite to the Right Angle, is equal to the Squares of the other two Sides taken together. For having de∣scribed the Squares of the other two Sides, AC dE, DE ab (taking ED=AB) and the Square of the greatest BC cb, it will be evident, that the parts X and Z are common to each, and that the two other Triangles in the greatest Square BAC and BDb, are equal to the two Triangles bac and Cdc which

remain in the two Squares of the lesser Sides; and so the whole truth of the Proposition will be evident, while these two things are undoubtedly true: 1. That the Side of the greatest Square Bb will necessarily concur with the Extremity of the less Db, and the other Side of the greatest Square Cc with its Extremity c, will precisely touch the Continuation of the Sides of the two least Squares dEa; as you'l see them both expressed in the Fi∣gure. 2. The said two Triangles are every way equal; for the Angles at C with the intermediate one at Z, make two Right ones, therefore they are equal; but the Side CA is equal to the Side cd, and CB to Cc, and the Angles at A and d Right ones. Wherefore if we conceive the Triangle ABC to to be turned about C, as a Center to the right hand, it will exactly agree with the Triangle Cdc, and the Point B will ne∣cessarily fall on the continued Line d E, as agreeing with the Line AB. Hence it is now evident, that Ca=BD, and be∣cause ba is also = bD, and the Angles at a and D Right ones. Where, if we conceive the Triangle bac to be moved about b as a Center, untill ba coincides with bD, and ac with DB, bc will also necessarily coincide with bB Q. E. D.

To this Demonstration of Van Schooten's, which we have thus illustrated and abbreviated, we will add another of our own, more like Euclids, but somewhat easier, which is this: Having drawn the Lines (as the other Figure 29 directs) the ▵ ACD being on the same Base AC with the Square AI, and between the same Parallels, is necessarily one half of it, but it is also half of the Parallelogram CF being on the same Base with it, viz. DC; therefore this Parallelogram = ▭ AI. In like manner ▵ ABE is half the ▭ AL, and also half the Parallelogram BF, therefore BF=▭AL: therefore CF + BF that is the ▭ of BD = to the two ▭ ▭ AI + AL. Q. E. D. For because the Side BE occurs to, or meets the Side LK, and the Side CD the Side IH continued, it yet more apparently follows; because the An∣gles a and b, and also c and d, are manifestly equal, as making both ways, with the Intermediate x or z, Right Angles. There∣fore the ▵ BAC being turned on the Center B and laid on BLE will exactly agree with it, and turned on the Center C and laid on CID, will agree with that also, &c.