Question 159047: Given: Parallelogram ABCD with Segment AB extended to E. Segment DFE intersects BC at F. Prove: AE/CD = AD/CF
Thanks!
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! you want to prove AE/CD = AD/CF
the key appears to be proving that triangles DCF and EBF are similar.
to prove that use AAA = AAA (all corresponding angles of one triangle are equal to all corresponding angles of the other triangle.
angle DFC = angle BFE (vertical angles created by intersecting lines are equal)
angle DCF = angle EBF (alternate interior angles of parallel lines are equal)
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.)
triangle DCF is similar to triangle EBF by AAA.
since all corresponding sides of similar triangle are proportional by the same ratio, CD/BE = CF/FB = DF/FE.
since this is a parallelogram, opposite sides are equal. this makes AD = CB and CD = AB
because of this equality, CD/BE becomes the same as AB/BE
you are asked to prove that AE/CD = AD/CF.
this becomes AE/AB = CB/CF because AB = CD and CB = AD.
we know that FB/CF = BE/CD because they are corresponding sides of similar triangles.
this equation becomes FB/CF = BE/AB because AB = CD
we want to prove that CB/CF = AE/AB
CB = CF + FB
AE = AB + BE
so CB/CF = AE/AB BECOMES
(CF+FB)/CF = (AB+BE)/AB
cross multiplying we get
(CF+FB)*AB = (AB+BE)*CF
multiplying out, we get
CF*AB + FB*AB = AB*CF + BE*CF
subtracting CF*AB from both sides of the equation, we get
FB*AB = BE*CF
dividing both sides of the equation by FB*BE, we get
AB/BE = CF/FB
which brings us back to the original equality that we know is good so the equation is valid.
this may be tough to see, but putting some numbers into an example may make it clearer.
if we know 1/2 = 3/6, then 1/(1+2) = 3/(3+6) should be valid.
1/3 = 3/9 is equal so the assumption is valid.
this allows us to say CB/CF = AE/AB
we know that AD = CB and CD = AB
so the equation becomes AD/CF = AE/CD which is what we originally tried to prove.'
i believe this proof is valid.
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).
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