The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

The 34. Theoreme. The 48. Proposition. If the square which is made of one of the sides of a triangle, be equall to the squares which are made of the two other sides of the same triangle: the angle comprehended vnder those two other sides is a right angle.

SVppose that ABC be a triangle, and let the square which is made of one of the sides there, namely, of the side BC, be equall to the squares which are made of the sides BA and AC. Then I say that the angle BAC is a right angle. Rayse vp (by the 11. propositiō) from the point A vnto the right line AC a perpendicular line AD. And (by the thirde proposition) vnto the line AB put an equall line AD. And by the first peticion draw a right line from the point D to the poin•• C. And forasmuch as

[illustration]

the line DA is equall to the line AB, the square which is made of the line DA is e∣quall to the square whiche is made of the line AB Put the square of the line AC, common to them both. VVherefore the squares of the lines DA and AC are equal to the squares of the lines BA and AC. But (by the proposition going before) the square of the line DC is equal to y^e squares of the lines. AD and AC. (For the angle DAC is a right angle) and the square of BC is (by supposition) equall to the squares of AB and AC. VVherefore the square of DC is equall to the square of BC: wherefore the side DC is equall to the side BC. And forasmuch as AB is equall to AD ••nd AC is common to them both, therefore these two sides DA and AC are equall to these two sides BA and AC, the one to the other, and the base DC is equall to the base BC•• wher∣fore (by the 8. proposition) the angle DAC is equall to the angle BAC. But the angle DAC is a right angle, wherefore also the angle BAC is a right angle. If

therefore the square which is made of one of the sides of a triangle, be equall to the squares which are made of the two other sides of the same triangle, the an∣gle comprehended vnder those two other sides is a right angle. which was re∣quired to be proued.

* 1.1This proposition is the conuerse of the former, and is of Pelitarius demon∣strated by an argument leading to an impossibilitie after this maner.

* 1.2Suppose that ABC be a triangle: & let the square of the side AC, be equal to the squares of the two sides AB and BC. Then I say that the angle at the point B, which is opposite to the side AC, is a right angle. For if the angle at the point

[illustration]

B be not a right angle, then shal it be eyther greater or lesse thē a right angle. First let it be is greater. And let the angle DBC be a right angle, by erecting from the point B a per∣pendicular line vnto the line BC (by the 11. proposition) which let be BD: and put the line BD equall to the lyne AB (by the thirde proposition). And drawe a line from C to D. Now (by the former proposition) the square of the side CD shalbe equall to the squares of the two sides BD and BC: wherefore also to the squares of the two sides BA and BC. Wherefore the base CD shalbe equall to the base CA, when as their squares are equall: which is con∣trary to the 24. proposition. For forasmuch as the angle ABC is greater then the angle DBC, and the two sides AB and BC are equall to the two sides DB and BC, the one to the other, the base CA shall be greater then the base CD. It is also contrary to the 7. proposition, for from the two endes of one & the same line, namely, frō the points B & C should be drawn on one and the same side two lines BD and DC ending at the pointe D, e∣quall to two other lines BA and AC drawen from the same endes and ending at an other point, namely, at A, which is impos∣sible. By the same reason also may we proue that the whole angle at the pointe B is not lesse then a right angle. Wherfore it is a right angle: which was required to be proued.

Notes

• * 1.1

  This proposition is the conue••se of the former.
• * 1.2

  An other De∣monstration af∣ter Pelitarius.