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

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Page 49

For , the

to the Triangle ACD . 20 . Conf.the

N. BHh Otherwise , Ii k K Having drawn a Diagonal and Perpendiculars , as ...

For , the

**Rect**. AIKC is equal to the Triangle ABC , and the**Rect**. ACLM is equalto the Triangle ACD . 20 . Conf.the

**Rect**.IKLM is equal to the Trap . ABCD , Ax.2 .N. BHh Otherwise , Ii k K Having drawn a Diagonal and Perpendiculars , as ...

Page 155

Through E and F , draw E G and FH , 3 parallel to AD and BC ; dividing the

D ABCD into three

both equal to Z ; -Con EG and FH are also equal Z. -- - Th . 15. 1. & Ax . 3 .

Through E and F , draw E G and FH , 3 parallel to AD and BC ; dividing the

**Rect**.D ABCD into three

**Rect**. AG , EH & FC . DEM . Now , because AD and BC areboth equal to Z ; -Con EG and FH are also equal Z. -- - Th . 15. 1. & Ax . 3 .

Page 166

Produce DA . Make EF equal EB , and AC equal to AF . Then is AB so divided , in

C , that ABR BC = ACO . Compleat the Squares ADHB and AFGC , and produce

GC to I. Dem . Then , because AD is bisected in E , and AF is added , the

Produce DA . Make EF equal EB , and AC equal to AF . Then is AB so divided , in

C , that ABR BC = ACO . Compleat the Squares ADHB and AFGC , and produce

GC to I. Dem . Then , because AD is bisected in E , and AF is added , the

**Rect**. Page 297

1 . if HE be added to both , the

under the two Extremes . Con . And , ADFE is a

Therefore , the Rectangle under the two Extremes , of four proportional Lines , is

...

1 . if HE be added to both , the

**Rect**. ADFE = AHIB - Ax.6.1 . But , AHIB is a**Rect**.under the two Extremes . Con . And , ADFE is a

**Rect**. under the two Means .Therefore , the Rectangle under the two Extremes , of four proportional Lines , is

...

Page 5

AEFD is not equal to the Area of the

AD , it will produce an Area equal to the

the Par . AEFD is equal to the

the ...

AEFD is not equal to the Area of the

**Rect**. ABCD . For , if AE be multiplied intoAD , it will produce an Area equal to the

**Rect**. ABCD , because , AB = AE . But ,the Par . AEFD is equal to the

**Rect**. AGHD only . For , it is demonstrable , that ,the ...

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A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ... Thomas Malton No preview available - 2016 |

### 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.