A Text-book of Euclid's Elements for the Use of Schools, Bücher 1Macmillan, 1904 - 456 Seiten |
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Seite 11
... perp . ,, perpendicular , pt . point ; and all obvious contractions of words , such as opp . , adj . , diag . , etc. , for opposite , adjacent , diagonal , etc. SECTION I PROPOSITION 1. PROBLEM . To describe an equilateral INTRODUCTORY .
... perp . ,, perpendicular , pt . point ; and all obvious contractions of words , such as opp . , adj . , diag . , etc. , for opposite , adjacent , diagonal , etc. SECTION I PROPOSITION 1. PROBLEM . To describe an equilateral INTRODUCTORY .
Seite 106
... perp . to PQ ; 1. 28 . ..the figures XH , HY are parms ; .. AH = XZ , and HK = ZY . I. 34 . But through C , the middle point of AB , a side of the △ ABK , CH has been drawn parallel to the side BK ; ... CH bisects AK : that is , AHHK ...
... perp . to PQ ; 1. 28 . ..the figures XH , HY are parms ; .. AH = XZ , and HK = ZY . I. 34 . But through C , the middle point of AB , a side of the △ ABK , CH has been drawn parallel to the side BK ; ... CH bisects AK : that is , AHHK ...
Seite 111
... perps drawn to the sides from X , Y , Z be concurrent . From Z and Y draw perps to AB , AC ; these perp3 , since they cannot be parallel , will meet at some point O. Because Join OX . Ax . 12 . B X It is required to prove that OX is perp ...
... perps drawn to the sides from X , Y , Z be concurrent . From Z and Y draw perps to AB , AC ; these perp3 , since they cannot be parallel , will meet at some point O. Because Join OX . Ax . 12 . B X It is required to prove that OX is perp ...
Seite 112
... perp . to the sides of the A. Then in the △ s OBP , OBR , Because { the also the OBP the L OBR , B P R OPB = the L ORB , being rt . L3 , and OB is common ; .. OP = OR . Similarly from the As OCP , OCQ , it may be shewn that OP = OQ ...
... perp . to the sides of the A. Then in the △ s OBP , OBR , Because { the also the OBP the L OBR , B P R OPB = the L ORB , being rt . L3 , and OB is common ; .. OP = OR . Similarly from the As OCP , OCQ , it may be shewn that OP = OQ ...
Seite 114
... perps AD , BE , CF be concurrent . Through A , B , and C draw straight lines MN , NL , LM parallel to the opposite ... perps to the sides of the △ LMN from their middle points . But these perps meet in a point : Ex . 3 , p . 60 . Ex . 1 ...
... perps AD , BE , CF be concurrent . Through A , B , and C draw straight lines MN , NL , LM parallel to the opposite ... perps to the sides of the △ LMN from their middle points . But these perps meet in a point : Ex . 3 , p . 60 . Ex . 1 ...
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Häufige Begriffe und Wortgruppen
ABCD AC is equal adjacent angles Algebra angle BAC angle equal base BC bisected bisectors centre chord circumference circumscribed circle concyclic Constr Describe a circle diagonal diameter divided equal angles equiangular Euclid Euclid's exterior angle find the locus given circle given point given straight line given triangle greater Hence hypotenuse inscribed circle isosceles triangle Let ABC line which joins magnitudes meet middle point nine-points circle opposite sides orthocentre par¹ parallelogram parm pass pedal triangle perp perpendiculars drawn plane XY polygon produced Proof proportional PROPOSITION PROPOSITION 13 prove quadrilateral radical axis radius rectangle contained rectilineal figure regular polygon right angles segment shew shewn side BC Similarly square straight line drawn tangent THEOREM triangle ABC twice the rect vertex vertical angle
Beliebte Passagen
Seite 353 - Pythagoras' theorem states that the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides.
Seite 340 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Seite 65 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Seite 162 - AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.
Seite 326 - From this it is manifest that prisms upon triangular bases, of the same altitude, are to one another as their bases. Let the...
Seite 162 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square...
Seite 291 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.
Seite 79 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Seite 18 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Seite 242 - We may here notice that the perpendiculars from the vertices of a triangle to the opposite sides are concurrent ; their meet is called the orthocentre, and the triangle obtained by joining the feet of the perpendiculars is called the pedal triangle.