Pure Mathematics, Volume 11874 |
From inside the book
Results 1-5 of 53
Page 92
... joining two of its opposite angles . All other four - sided figures are called trapeziums . Postulates . 1. Let it be granted that a straight line may be drawn from any one point to any other point . 2. That a terminated straight line ...
... joining two of its opposite angles . All other four - sided figures are called trapeziums . Postulates . 1. Let it be granted that a straight line may be drawn from any one point to any other point . 2. That a terminated straight line ...
Page 98
... Join FC , GB . PROOF . - Because AF is equal to AG ( Construction ) , and AB is equal to AC ( Hyp . ) , Therefore the two sides FA , AC are equal to the two sides GA , AB , each to each ; And they contain the angle FAG , common to the ...
... Join FC , GB . PROOF . - Because AF is equal to AG ( Construction ) , and AB is equal to AC ( Hyp . ) , Therefore the two sides FA , AC are equal to the two sides GA , AB , each to each ; And they contain the angle FAG , common to the ...
Page 99
... Join DC . PROOF .-- Because in the triangles DBC , ACB , DB is equal to AC , and BC is common to both , Therefore the two sides DB , BC are equal to the two sides AC , CB , each to each ; And the angle DBC is equal to the angle АСВ ...
... Join DC . PROOF .-- Because in the triangles DBC , ACB , DB is equal to AC , and BC is common to both , Therefore the two sides DB , BC are equal to the two sides AC , CB , each to each ; And the angle DBC is equal to the angle АСВ ...
Page 100
... Join CD . PROOF . - Because AC is equal to AD ( Hyp . ) , The triangle ADC is an isosceles triangle , and the angle ACD is therefore equal to the angle ADC ( I. 5 ) . But the angle ACD is greater than the angle BCD ( Ax . 9 ) ...
... Join CD . PROOF . - Because AC is equal to AD ( Hyp . ) , The triangle ADC is an isosceles triangle , and the angle ACD is therefore equal to the angle ADC ( I. 5 ) . But the angle ACD is greater than the angle BCD ( Ax . 9 ) ...
Page 102
... Join DE . Upon DE , on the side remote from A , de- scribe an equilateral triangle DEF ( I. 1 ) . Join AF . Then the straight line AF shall bisect the angle BAC . PROOF . - Because AD is equal to AE ( Const . ) , and AF is common to the ...
... Join DE . Upon DE , on the side remote from A , de- scribe an equilateral triangle DEF ( I. 1 ) . Join AF . Then the straight line AF shall bisect the angle BAC . PROOF . - Because AD is equal to AE ( Const . ) , and AF is common to the ...
Other editions - View all
Common terms and phrases
a²b a²x² ab² ABCD algebraical angle ABC angle ACB angle BAC base BC BC is equal bisect brackets cent centim centre circle ABC circumference coefficient common Const cos² cosec cube root decimal figures denominator divided divisor equation expression exterior angle factor Find the value fraction given straight line gnomon gram greater Hence integer join kilom less Let ABC logarithm metres miles millig millim Multiply opposite angles parallel parallelogram perpendicular PROOF.-Because Q. E. D. Proposition quotient ratio rectangle contained remainder right angles segment sides sine square on AC square root subtraction touches the circle triangle ABC twice the rectangle unknown quantity x²y x²y² xy³