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Prove the following. If B is the midpoint of line AC, D is the midpoint of line CE and line AB is congruent to line DE, then line AE = 4AB

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Something is missing –  i. m. soloveichik Sep 5 '12 at 0:28
Can you do an informal proof? Is the problem formalizing a proof or finding one in the first place? What have you tried? –  Ross Millikan Sep 5 '12 at 0:34
Note that this fails in $\mathbb R^2$. Even in $\mathbb R$, unless the distinct labels are known to be distinct points, it is not true. Try $A=E=0, B=D=1, C=2$ –  Ross Millikan Sep 5 '12 at 0:37
It is highly likely (based on Ross's second comment alone!) that at least one of the following has occurred: (1) You there are additional conditions/assumptions that you've not shared; (2) The problem is quite simply wrong; (3) We're all missing/misinterpreting something confusing in the problem. –  Cameron Buie Sep 5 '12 at 0:45
I've only ever heard the phrase "two-column proof" in a first (high school) course in plane geometry, so I'd expect a picture would accompany the problem statement to fill out the assumptions, a la Euclid on his bad days. Probably ABCDE are all distinct and collinear. –  Kevin Carlson Sep 5 '12 at 1:26

1 Answer 1

Well, I'm assuming you are talking about a line segment from $A$ to $E$ where all points are distinct (I'm guessing an image went with this question that you didn't post).

$\underline{Statement}$ $\hspace 3in \underline{Reason}$

$1)\overline{AB} \cong \overline{DE} \hspace 2.85in 1)$ Given

$2)B$ and $D$ are midpoints of $\overline{AC}$ and $\overline{CE} \hspace .89in 2)$ Given

$3) AE=AB+BC+CD+DE \hskip 1.54in 3)$ Given

$4)\overline{AB} \cong \overline{BC}$ and $\overline{CD} \cong \overline{DE} \hskip 1.74in 4)$ Property of Midpoint

$5) \overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DE} \hskip 1.88in 5)$ Transitive Property using $1)$ and $4)$

$6)AE=4AB \hskip 2.78in 6)$ Substitution using $3)$ and $5)$

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