## A Supplement to the Elements of Euclid |

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Page 3

... a perpendicular to it . Let XY be a given straight line indefinite in length , and A , B , two

... a perpendicular to it . Let XY be a given straight line indefinite in length , and A , B , two

**given points**without it ; not - situated in a perpendicular to XY : It is required to find**a point**in XY that shall be equidistant from A and B. Page 4

... point D is equidistant from A , B. ACD , For , join A , D and B , D. Then , since ( constr . ) AC BC , and CD is common to the two BCD , and that ( constr . and E. 10. def . 1. ) = 2 ...

... point D is equidistant from A , B. ACD , For , join A , D and B , D. Then , since ( constr . ) AC BC , and CD is common to the two BCD , and that ( constr . and E. 10. def . 1. ) = 2 ...

**given points**without that line 4 A SUPPLEMENT TO THE. Page 5

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**point**in an indefinite straight line DZ , which bisects the**given**finite straight line AB , at right angles , is equidistant from the extremi- ties**A**and B , of that**given**finite line : And , any**point**which is not in that indefinite ... Page 6

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**a given**triangle be bisected , the perpendiculars drawn to the sides , from the three several bisections , shall all meet in the same**point**: And that**point**is equidistant from the three angular**points**of the**given**triangle . Let ABC be**a**... Page 7

... point G : And , since it has been shown that AG = BG = CG , the point G is equidistant from A , B and C. PROP . V. 9. PROBLEM . To find

... point G : And , since it has been shown that AG = BG = CG , the point G is equidistant from A , B and C. PROP . V. 9. PROBLEM . To find

**a point**, in a given plane , which shall be equidistant from three**given points**in the plane , that ...### Other editions - View all

### Common terms and phrases

ABCD bisect centre chord circle ABC circle described circumference constr decagon describe a circle describe the circle diameter distance divided double draw a straight draw E equi equiangular equilateral and equiangular EUCLID Euclid's Elements F draw find a point finite straight line given circle given finite straight given point given ratio given square given straight line half hypotenuse inscribed isosceles join K less Let ABC lines be drawn manifest manner meet the circumference number of equal number of sides parallel to BC parallelogram pass perimeter point G polygon PROBLEM produced PROP rectangle contained rectilineal figure remaining sides required to describe required to draw rhombus right angles segment semi-diameter straight line cutting straight line joining tangent THEOREM three given touch the circle trapezium vertex

### Popular passages

Page 278 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.

Page 326 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the...

Page 76 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Page 2 - ... angles equal; and conversely if two angles of a triangle are equal, two of the sides are equal. 3. If two triangles have the three sides of one equal to the three sides of the other, each to each, do you think the two triangles are alike in every respect ? 4. If two triangles have the three angles of one equal to the three angles of the other, each to each, do you think the two triangles are necessarily alike in every respect ? 5. Draw two triangles, the angles of one being equal to the angles...

Page 309 - Divide a straight line into two parts such that the rectangle contained by the whole line and one of the parts shall be equal to the square on the other part.

Page 256 - ... line and the extremities of the base have the same ratio which the other sides of the triangle have to one...

Page 175 - 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...

Page 327 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Page 86 - In every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular...

Page 5 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC. In BC take any point D, and join AD; and at the point A, in the straight line AD, make (I.