## Plane GeometryMacmillan, 1901 |

### From inside the book

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Page 16

... equation of the locus of a

... equation of the locus of a

**point equidistant**from the points (a1, b1) and (ag, b2), is (a1 – a 2) & + (b1 – bo) y + c = 0 then the value of 'C' is : (b). (aft. –a4+bi-. b%. (d) N(aī + bi – ań – b%) Aline cutting offintercept—3 from the y ... Page 11

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**point equidistant**from A and B. Draw the figure to scale and find approxi- w mately the number of feet of pipe that will be required to convey the water from ... Distance of a Point from a Line . Suppose**POINTS EQUIDISTANT**FROM TWO POINTS 11. Page 22

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**point equidistant**from two given points is the perpendicular bisector of the line joining the points. Theorems on Locus : (a) The locus of a**point equidistant**from a fixed point is a circle with the fixed point as centre. (b) The locus ... Page 22

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**point equidistant**from the extremeties of a straight line is the perpen- dicular erected at the middle of the line . ( See Nos . 44 , 43. ) 84 . Theorem . - Every point in the bisector of an angle is equidistant from the sides of the ... Page 223

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**points**not in a straight line . To prove that one circle , and one only , can be drawn to pass through A , B and C. Proof It is only necessary to show that there is one**point**( and one only )**equidistant**from A , B , and C. Now the ...### Other editions - View all

### Common terms and phrases

AABC ABCD adjacent angles algebraic altitude angle equal angle formed angles are equal apothem base angle bisect bisector central angle circumference construct a triangle decagon diagonals diagram for Prop diameter draw drawn equiangular equiangular polygon equilateral triangle equivalent exterior angle find a point Find the area given circle given line given point given triangle HINT homologous sides hypotenuse inscribed isosceles triangle joining the midpoints line joining mean proportional measured by arc median opposite sides parallel lines parallelogram perimeter perpendicular perpendicular-bisector point equidistant produced proof is left PROPOSITION prove Proof proving the equality quadrilateral radii rectangle regular hexagon regular polygon rhombus right angle right triangle SCHOLIUM School secant segments side equal similar polygons similar triangles straight angle straight line tangent THEOREM third side transversal trapezoid triangle ABC triangle are equal vertex vertical angle

### Popular passages

Page 148 - If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other.

Page 180 - 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. B' ADC A' D' C' Hyp. In triangles ABC and A'B'C', ZA = ZA'.

Page 45 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second.

Page 31 - The median to the base of an isosceles triangle is perpendicular to the base.

Page 209 - The area of a regular polygon is equal to onehalf the product of its apothem and perimeter.

Page 130 - If a line divides two sides of a triangle proportionally, it is parallel to the third side.

Page 71 - The midpoints of two opposite sides of a quadrilateral and the midpoints of the diagonals determine the vertices of a parallelogram. * Ex.

Page 26 - If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles triangles.

Page 190 - The areas of two similar triangles are to each other as the squares of any two homologous sides.

Page 56 - A line which joins the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.