document.write( "Question 159047: Given: Parallelogram ABCD with Segment AB extended to E. Segment DFE intersects BC at F. Prove: AE/CD = AD/CF
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Algebra.Com's Answer #117197 by gonzo(654)\"\" \"About 
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you want to prove AE/CD = AD/CF
\n" ); document.write( "the key appears to be proving that triangles DCF and EBF are similar.
\n" ); document.write( "to prove that use AAA = AAA (all corresponding angles of one triangle are equal to all corresponding angles of the other triangle.
\n" ); document.write( "angle DFC = angle BFE (vertical angles created by intersecting lines are equal)
\n" ); document.write( "angle DCF = angle EBF (alternate interior angles of parallel lines are equal)
\n" ); document.write( "angle CDF = angle FEB (if 2 angles of a triangle equal 2 angles of another triangle then the 3d angles must be equal since the sum of the interior angles of all triangles equals 180.)
\n" ); document.write( "triangle DCF is similar to triangle EBF by AAA.
\n" ); document.write( "since all corresponding sides of similar triangle are proportional by the same ratio, CD/BE = CF/FB = DF/FE.
\n" ); document.write( "since this is a parallelogram, opposite sides are equal. this makes AD = CB and CD = AB
\n" ); document.write( "because of this equality, CD/BE becomes the same as AB/BE
\n" ); document.write( "you are asked to prove that AE/CD = AD/CF.
\n" ); document.write( "this becomes AE/AB = CB/CF because AB = CD and CB = AD.
\n" ); document.write( "we know that FB/CF = BE/CD because they are corresponding sides of similar triangles.
\n" ); document.write( "this equation becomes FB/CF = BE/AB because AB = CD
\n" ); document.write( "we want to prove that CB/CF = AE/AB
\n" ); document.write( "CB = CF + FB
\n" ); document.write( "AE = AB + BE
\n" ); document.write( "so CB/CF = AE/AB BECOMES
\n" ); document.write( "(CF+FB)/CF = (AB+BE)/AB
\n" ); document.write( "cross multiplying we get
\n" ); document.write( "(CF+FB)*AB = (AB+BE)*CF
\n" ); document.write( "multiplying out, we get
\n" ); document.write( "CF*AB + FB*AB = AB*CF + BE*CF
\n" ); document.write( "subtracting CF*AB from both sides of the equation, we get
\n" ); document.write( "FB*AB = BE*CF
\n" ); document.write( "dividing both sides of the equation by FB*BE, we get
\n" ); document.write( "AB/BE = CF/FB
\n" ); document.write( "which brings us back to the original equality that we know is good so the equation is valid.
\n" ); document.write( "this may be tough to see, but putting some numbers into an example may make it clearer.
\n" ); document.write( "if we know 1/2 = 3/6, then 1/(1+2) = 3/(3+6) should be valid.
\n" ); document.write( "1/3 = 3/9 is equal so the assumption is valid.
\n" ); document.write( "this allows us to say CB/CF = AE/AB
\n" ); document.write( "we know that AD = CB and CD = AB
\n" ); document.write( "so the equation becomes AD/CF = AE/CD which is what we originally tried to prove.'
\n" ); document.write( "i believe this proof is valid.
\n" ); document.write( "the keys are similar triangles having corresponding sides proportional, opposite sides of parallelograms being equal, and the fact that if a/b = c/d then a/(a+b) = c/(c+d).\r
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