Inductive Plane Geometry With Numerous Exercises Theorems And Problems for Ad
Inductive Plane Geometry With Numerous Exercises Theorems And Problems for Ad
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To quickly assess the difficulty of the text, read a short excerpt:
(?) . -. GJf bisects CD. (?) Similarly, LH is II to AF and bisects CD.
. -. LH, MO, and ED are concurrent in point K.
E-egarding AE as a transversal of the triangle CDF, AD ' BF- EC= AF ' BC ' ED. (?) AD BF EC^AF BCED^ /^s '** 2 * 2 * 2 2 ' 2 ' 2 * ^'^ . -. LK' MO' NH=LH - MK • iV^(7. (?) . -. L, M, and N are collinear. (?) q. E. D.
If lines drawn from the vertices of a triangle are con- current, the product of three non-adjacent segments of the sides equals the product of the other three.
Post.... Let ABC be a triangle, any point in its plane through which pass the lines AQ, CN, and BH.
To Prove. BQ - HC - AN= QC - AH - NB.
Dem. BQ. -HC-AO = BC'AH' QO.
(BH is a transversal cutting A ACQ. ) qC ' AG ' NB==BC ' QO ' NA.
(JSfC is a transversal cutting A ABQ. ) 172 Plane Geometry.
" qC'NB NA ^'^ . -. BQ'HG'NA^qC^NB' AH, (?) q. E. D.
(This is Ceva's theorem, discovered in 1678, and the follow- ing is its converse. ) 987. If three points are given on the sides of a triangle, or the sides extended so that the product of three non-adjacent segments of the sides equals the product of the other three, the lines joining these points with the opposite vertices are concurrent.
What to read after Inductive Plane Geometry With Numerous Exercises Theorems And Problems for Ad?
You can find similar books in the "Read Also" column, or choose other free books by G Irving George Irving Hopkins to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
(?) . -. GJf bisects CD. (?) Similarly, LH is II to AF and bisects CD.
. -. LH, MO, and ED are concurrent in point K.
E-egarding AE as a transversal of the triangle CDF, AD ' BF- EC= AF ' BC ' ED. (?) AD BF EC^AF BCED^ /^s '** 2 * 2 * 2 2 ' 2 ' 2 * ^'^ . -. LK' MO' NH=LH - MK • iV^(7. (?) . -. L, M, and N are collinear. (?) q. E. D.
If lines drawn from the vertices of a triangle are con- current, the product of three non-adjacent segments of the sides equals the product of the other three.
Post.... Let ABC be a triangle, any point in its plane through which pass the lines AQ, CN, and BH.
To Prove. BQ - HC - AN= QC - AH - NB.
Dem. BQ. -HC-AO = BC'AH' QO.
(BH is a transversal cutting A ACQ. ) qC ' AG ' NB==BC ' QO ' NA.
(JSfC is a transversal cutting A ABQ. ) 172 Plane Geometry.
" qC'NB NA ^'^ . -. BQ'HG'NA^qC^NB' AH, (?) q. E. D.
(This is Ceva's theorem, discovered in 1678, and the follow- ing is its converse. ) 987. If three points are given on the sides of a triangle, or the sides extended so that the product of three non-adjacent segments of the sides equals the product of the other three, the lines joining these points with the opposite vertices are concurrent.
What to read after Inductive Plane Geometry With Numerous Exercises Theorems And Problems for Ad?
You can find similar books in the "Read Also" column, or choose other free books by G Irving George Irving Hopkins to read onlineMoreLess
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