The elements of Euclid for the use of schools and colleges : comprising the first two books and portions of the eleventh and twelfth books; with notes and exercises . e square described on the side BG shall beequal to the squares described on the sides BA, AC. On BG describethe square BDEG,and on BA, AG de-scribe the squaresGB,HG, [ A draw ALparallel to BD orCE; [ andjoin^i), FC Then, because theangle BAG is a rightangle, {Hypothesis. and that the angleBAG is also a rightangle, {Definition 30. the two straight lines AG, AG, on tiie opposite sides ofAB, make with it at the poin
The elements of Euclid for the use of schools and colleges : comprising the first two books and portions of the eleventh and twelfth books; with notes and exercises . e square described on the side BG shall beequal to the squares described on the sides BA, AC. On BG describethe square BDEG,and on BA, AG de-scribe the squaresGB,HG, [ A draw ALparallel to BD orCE; [ andjoin^i), FC Then, because theangle BAG is a rightangle, {Hypothesis. and that the angleBAG is also a rightangle, {Definition 30. the two straight lines AG, AG, on tiie opposite sides ofAB, make with it at the point A the adjacent angles ecjualto two right angles ; therefore GA is in the same straight line mth AG. [I. the same reason, AB and AH Sire in the same straightline. Now the angle DBG is equal to the angle FBA, for eachof them is a right angle. [Axiora 11. Add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angleFBC [Axiom 2. And because the two sides AB, BD are equal to the twosides FB, BG, each to each ; [Definition 30. and the angle DBA is equal to the angle FBC; therefore the triangle ABD la equal to the triangle^^O. ^ [ BOOK I. 47, 48. 51 / Now the parallelo^ani BL is double of the triangleABD, because they are on the same ba^e BD, and betweenthe same parallels BD, AL. [I. 41. And the square GB is double of the triangle FBC, because4they are on the same base FB. and between tlie sameparallels FB, GO. [I. 41. But the doubles of equals are equal to one another. [Ax. the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can beshewn, that the parallelogram CL is equal to the square the whole square BDEC is equal to the twosquares GB, H(\ [Axiovi 2. And the square BDECis described on BC\ and tlie squaresGB, HC on BA, AC. Therefore the square described on the side BC is equal tothe squares described on the sides BA, AC. Wherefore, in any right-angled triangle &c. PROPOSITION
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