Elements of geometry and trigonometry . B D PROPOSITION IV. THEOREM. If a straight line he perpendicular to two straight lines at theirpoint of intersection, it will he perpendicular to the plane ofthose Let MN be the plane of thetwo Unes BB, CC, and let APbe perpendicular to them attheir point of intersection P ;then will AP be perpendicularto every line of the plane pass-ing through P, and corsequentlyto the plane itself (Def. 1.). Through P, draw in the planeMN, any straight line as PQ,and through any point of thisline, as Q, drawBQC, bo that BQ shall be equal to QC (BookIV. Prob. V
Elements of geometry and trigonometry . B D PROPOSITION IV. THEOREM. If a straight line he perpendicular to two straight lines at theirpoint of intersection, it will he perpendicular to the plane ofthose Let MN be the plane of thetwo Unes BB, CC, and let APbe perpendicular to them attheir point of intersection P ;then will AP be perpendicularto every line of the plane pass-ing through P, and corsequentlyto the plane itself (Def. 1.). Through P, draw in the planeMN, any straight line as PQ,and through any point of thisline, as Q, drawBQC, bo that BQ shall be equal to QC (BookIV. Prob. V.) ; draw AB, AQ, AC. The base BC being divided into two equal parts at the pointQ, the triangle BPC will give (Book IV. Prop. XIV.),PCHPB2=2PQH2QC. The triangle BAC will in like manner give,AC2+AB-=2AQ^ + 2QC2. Taking the first equation from the second, and observingthat the triangles APC, APB, which are both right angled atP, give AC2_PC2^AP2, and AB^—PB^^AP^;we shall have AP2+AP2=2AQ2—2PQ2. Therefore, by taking the halves of both, we haveAP^-AQ^—PQ^ or AQ2=.APHPQ ;hence the t?iianrrlc APQ is right angled at P ; hence AP is per-pendicuL^r to PQ. BOOK VI. 129 Scholium. Thus it is evident, not only
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