Geometrical Exercises for BeginnersMacmillan, 1882 - 203 sider |
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Side 20
... second case , when the angles PBA , ACD are equal to each of these equals add the angle ABD , when the sum of the angles PBA , ABD is equal to the sum of the angles ABD , ACD ; but the former sum is ( I. 13 ) equal to two right angles ...
... second case , when the angles PBA , ACD are equal to each of these equals add the angle ABD , when the sum of the angles PBA , ABD is equal to the sum of the angles ABD , ACD ; but the former sum is ( I. 13 ) equal to two right angles ...
Side 23
... BIL G is circumscribable by a circle . But the angle BIL is a right angle , therefore ( III . 22 ) the angle AGB is a right angle , in other words , AG is perpendicular to BC . To prove the second part . Join IG , HG FOR BEGINNERS . 23.
... BIL G is circumscribable by a circle . But the angle BIL is a right angle , therefore ( III . 22 ) the angle AGB is a right angle , in other words , AG is perpendicular to BC . To prove the second part . Join IG , HG FOR BEGINNERS . 23.
Side 24
... second part . EXAMPLES . 1. Prove that the point , L , in the preceding proposition , is the centre of the inscribed circle of the triangle GHI . 2. Prove that the sides BC , CA , AB of the triangle ABC are the external bisectors of the ...
... second part . EXAMPLES . 1. Prove that the point , L , in the preceding proposition , is the centre of the inscribed circle of the triangle GHI . 2. Prove that the sides BC , CA , AB of the triangle ABC are the external bisectors of the ...
Side 73
... second case . Draw PT perpendicular to LM . Divide PT in S , so that PS : PT in the given ratio ( VI . 10 ) . Draw SK parallel to LM , and produce it indefinitely : this is the required locus . For , evidently ( VI . 4 ) , every ...
... second case . Draw PT perpendicular to LM . Divide PT in S , so that PS : PT in the given ratio ( VI . 10 ) . Draw SK parallel to LM , and produce it indefinitely : this is the required locus . For , evidently ( VI . 4 ) , every ...
Side 127
... second equation by n , and adding , we have mAC2 + nBC2 = mAO2 + nBO2 + ( m + n ) CO2 + 20D ( mAO - nBO ) . But m40 obviously equals nBO , since BO contains m , and AO n equal parts ; therefore the last term of the equation , viz . 20D ...
... second equation by n , and adding , we have mAC2 + nBC2 = mAO2 + nBO2 + ( m + n ) CO2 + 20D ( mAO - nBO ) . But m40 obviously equals nBO , since BO contains m , and AO n equal parts ; therefore the last term of the equation , viz . 20D ...
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Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Vanlige uttrykk og setninger
AD² angle BAC angle BCD angle equal Assistant-Master base angles base BC BD² Bisect Cambridge centre chord circumference circumscribing circle Clifton College Conic Sections Consider Prop construct Crown 8vo draw ELEMENTARY TREATISE English equal to half escribed circle Eton College Extra fcap Fellow of St find the locus fixed point GEOMETRY given base given circle given difference given ratio given square given straight line Given the base given vertical angle GRAMMAR GREEK half a right half the given Hence HISTORY hypotenuse inscribed internal bisector intersection isosceles J. P. MAHAFFY John's College late Fellow LATIN Let ABC Mathematical meet middle point nine-point circle numerous Illustrations Owens College Oxford parallel perpendicular perpendicular to BC Professor prove quadrilateral R. C. JEBB radius rectangle rectangle contained right angle right-angled triangle School segment semicircle tangent triangle ABC Trinity College vertex W. K. CLIFFORD
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