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If ab plus bc equals ac then ac equals ab plus bc?
yes because ab plus bc is ac
Factor 6ab minus 6ac plus 6b squared minus 6bc?
6(ab - ac + b2 - bc)
The vertex of ABC is?
there will be three vertex AB, BC, AC
How do you prove that if the greatest common factor of A and B equals 1 and C does not equal 0 then the greatest common factor of AC and BC divides C?
If the greatest common factor/divisor of A and B is 1 then they are coprime - they do not share any prime factors. Multiplying both through by C means, obviously, that each number now divides by C. In fact, C is their greatest common divisor, since AC and BC do not have further common factors after C is taken out. Hence the GCF of AC and BC is not merely a factor of C - it is C. (The question makes sense only if A, B and C are integers.)
Prove the mid-line theorem?
Let ABC be a triangle. Let D and E be the mid points of AB and AC respectively. Then the mid-line theorem states that DEBC and DE = BC/2.Extend DE beyond E to F such that DE = EF. Since AE = CE, triangles ADE and CEF are equal, making CFAB (or CFBD, which is the same) because, for the transversal AC, the alternating angles DAE and ECF are equal. Also,CF = AD = BD, such that BDFC is a parallelogram. It follows that BC = DF = 2·DE which is what we set out to prove.Conversely, let D be on AB, E on AC, DEBC and DE = BC/2. Prove that AD = DB and AE = CE.This is because the condition DEBC makes triangles ADE and ABC similar, with implied proportion,AB/AD = AC/AE = BC/DE = 2.It thus follows that AB is twice as long as AD so that D is the midpoint of AB; similarly, E is the midpoint of AC.
