Elements of Geometry and Trigonometry |
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Page 80
... consequently BD2 = BC2 + CD2—2BC × CD ( Prop . IX . ) . Adding AD to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 AC2 , we have AB2 = BC2 + AC2-2BC × CD . - Secondly . When the ...
... consequently BD2 = BC2 + CD2—2BC × CD ( Prop . IX . ) . Adding AD to each , and observing that the right angled trian- gles ABD , ADC , give AD2 + BD2 = AB2 , and AD2 + CD2 AC2 , we have AB2 = BC2 + AC2-2BC × CD . - Secondly . When the ...
Page 92
... consequently similar ( Prop . XX . ) . In the same manner it might be shown that all the remaining triangles are similar , whatever be the number of sides in the polygons pro- posed : therefore two similar polygons are composed of the ...
... consequently similar ( Prop . XX . ) . In the same manner it might be shown that all the remaining triangles are similar , whatever be the number of sides in the polygons pro- posed : therefore two similar polygons are composed of the ...
Page 210
... consequently equal to , the first ; hence , or x2 √3 = ax + bx + cx = x ( a + b + c ) ; x√3 = a + b + c a + b + c therefore , x = √3 REMARK . Since the perpendicular CH is equal to x√3 , it is consequently equal to a + b + c : that ...
... consequently equal to , the first ; hence , or x2 √3 = ax + bx + cx = x ( a + b + c ) ; x√3 = a + b + c a + b + c therefore , x = √3 REMARK . Since the perpendicular CH is equal to x√3 , it is consequently equal to a + b + c : that ...
Contents
BOOK | 7 |
Problems relating to the First and Third Books 57 | 57 |
BOOK IV | 68 |
14 other sections not shown
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone consequently convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM Prop proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant segment similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line TABLE OF LOGARITHMIC tang tangent THEOREM triangle ABC triangular prism vertex