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If A, B, C, D, and E are collinear with AC=10, B between A and C, C between B and D, and D between C and E, and AC=BD=CE, then AB=BC=DE

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Final answer:

Given the collinear points and segment equality, it is proven that the lengths AB, BC, and DE are equal when AC=BD=CE and they all equal 10.

Step-by-step explanation:

If A, B, C, D, and E are collinear with AC=10, B is between A and C, C is between B and D, and D is between C and E, it's given that AC=BD=CE. From this information, we want to determine the lengths of AB, BC, and DE.

Since AC is the distance from A to C and AC=10, and it's given that AC=BD=CE, we have that BD=CE=10 as well. Now, because B, C, and D are collinear in that order, BD is made up of AB and BC, so AB + BC = BD. Similarly, CE is composed of CD and DE, so CD + DE = CE. Given that CD is the same length as BC because they are both between the same two segments (AC and BD) which are equal in length, we have BC = CD.

So, we can say that AB + BC = 10 and BC + DE = 10. However, since BC = CD, and CD and DE are part of the same segment CE which equals 10, that means BC = DE. With BC as a common length in both equations, we can conclude that AB equals DE.

Therefore, we have shown that AB = BC = DE based on the given collinear relationships and segment equality.

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