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In the figure below C is the mid-point of segment AE and AB < DE. Which is more (Ans A)



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I know its going to be like $B-A < E-D$ . How will i incorporate BC and CD into the above equation ?

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$AB+BC=CD+DE\Rightarrow BC-CD=DE-AB$. Note $DE-AB$ is positive. –  David Mitra Aug 4 '12 at 1:07

2 Answers 2

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$$BC=AC-AB$$ $$CD=CE-DE$$

Now use that $\,AC=CE\,,\,\,AB<DE\,$ and...voilá!

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Draw your own number line, carefully, with $C$ exactly halfway between $A$ and $E$, and $AB$ significantly less than $DE$. Can you see that $BC$ is bigger than $CD$?

If not, experiment with numbers. Suppose that the distance between $A$ and $C$ is $10$, as is the distance between $C$ and $E$. Suppose that $AB=2$ and $DE=4$. Which is bigger, $BC$ or $CD$?

Let's put it another way. We have $5$ people who live along a long straight road. Person $C$ lives exactly halfway between $A$ and $E$. We are given that $B$ is closer to $A$ than $D$ is to $E$. Can you see that $B$ is farther from $C$ than $D$ is?

Or else we can use "algebra." We have $$AB+BC=CD+DE,\tag{$1$}$$ because $C$ is halfway between $A$ and $E$. We are told that $AB<DE$. That means that $DE-AB>\gt 0$.

From Equation $(1)$, we get $$DE-AB=BC-CD.$$ Since $DE-AB \gt 0$, we have $BC-CD\gt 0$. Thus $BC\gt CD$.

Remark: It seems to me that what the problems you have been posting have in common is that you are expected to use numerical and geometric common sense, and usually little or no algebra. So probably you are expected to reach the answer by an argument like the ones described at the beginning of the answer, and not by manipulation of inequalities.

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