## A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the Mathematics. ... By Thomas Malton. ... |

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Page 68

А Y B X N But , if the Proportional was required to be to Z , as Y to X ; then the

Rectangle , or any Parallelogram , ABCD , must be under the two Lines Y and Z ;

which will , in this Cafe , be the two

now ...

А Y B X N But , if the Proportional was required to be to Z , as Y to X ; then the

Rectangle , or any Parallelogram , ABCD , must be under the two Lines Y and Z ;

which will , in this Cafe , be the two

**Means**; and DG the Proportional fought , isnow ...

Page 291

I. In right angled Triangles , the Perpendicular ( BD ) is a

between the Segments ( AD & DC ) of the Hypothenuse , made by the

Perpendicular . For , the Triangles ABD , DBC , being similar , the Side AD ( of the

Triangle ABD ) ...

I. In right angled Triangles , the Perpendicular ( BD ) is a

**mean**Proportionalbetween the Segments ( AD & DC ) of the Hypothenuse , made by the

Perpendicular . For , the Triangles ABD , DBC , being similar , the Side AD ( of the

Triangle ABD ) ...

Page 330

The Diameter or Perpendicular , of a regular Pentagon , is divided in extreme and

ABCDE , let the Diameter C F be drawn , and the Diagonal G HD BD , cuting C F

in ...

The Diameter or Perpendicular , of a regular Pentagon , is divided in extreme and

**mean**Proportions by ' a Diagonal cuting it at Right Angles . In the PentagonABCDE , let the Diameter C F be drawn , and the Diagonal G HD BD , cuting C F

in ...

Page 403

Find X a

and C I added to LL . Th . II . Dem . Now , AC , CE , EG , & GB are equal , by

Construction . And , ABXBC = ACXCD + EF - GH ( Th.13 . ) i . e . to ACXC1 , +

CEXCI + ...

Find X a

**mean**Proportional between AC and CI ; also Y , a**Mean**, between CEand C I added to LL . Th . II . Dem . Now , AC , CE , EG , & GB are equal , by

Construction . And , ABXBC = ACXCD + EF - GH ( Th.13 . ) i . e . to ACXC1 , +

CEXCI + ...

Page 405

But , a Circle , whose Radius is equal to a

between AB and BO , is equal to all the conical Superficies , described by the

revolution of the Poligon , whose Sides are equal to AF or AO - Th.14 .

Consequently , a ...

But , a Circle , whose Radius is equal to a

**Mean**, between AB and BF , orbetween AB and BO , is equal to all the conical Superficies , described by the

revolution of the Poligon , whose Sides are equal to AF or AO - Th.14 .

Consequently , a ...

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### Common terms and phrases

ABCD added alſo Altitudes analogous Area Baſe becauſe biſected Book called Center Chord Circle Circumference common Cone conf conſequently Conſtruction contained cuting Cylinder Demonſtration deſcribe Diagonal Diameter difference divided draw drawn equal Euclid evident extreme fame Feet Figure firſt formed four fourth given given Line greater half Hence Inches inſcribed join laſt leſs manner mean meaſure multiplied muſt oppoſite parallel Parallelogram Parallelopiped Pentagon perpendicular Plane Point Poligon Priſm Prob PROBLEM produced Proportion Propoſition proved Pyramid Quantities Radius Ratio Rect Rectangle reſpectively Right Angles Right Line ſame ſame Ratio ſay ſeeing Segment Sides ſimilar Solid ſome Sphere Square ſuch Surface taken Terms THEOREM third thoſe touch Triangle uſe wherefore whole whoſe

### Popular passages

Page 124 - When you have proved that the three angles of every triangle are equal to two right angles...

Page 221 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Page 285 - EG, let fall from a point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter DS, EF (p.

Page 284 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.

Page 186 - From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles.

Page 248 - To express that the ratio of A to B is equal to the ratio of C to D, we write the quantities thus : A : B : : C : D; and read, A is to B as C to D.

Page 161 - In any triangle, if a line be drawn from the vertex at right angles to the base; the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.

Page 160 - In any isosceles triangle, the square of one of the equal sides is equal to the square of any straight line drawn from the vertex to the base plus the product of the segments of the base.

Page 250 - Ratios that are the same to the same ratio, are the same to one another. Let A be to B as C is to D ; and as C to D, so let E be to F.

Page 124 - Angles, taken together, is equal to Twice as many Right Angles, wanting four, as the Figure has Sides.