# Isosceles Trapezium problem

I came across a problem in a certain quiz which I couldn't solve.

Here it is reproduced:

Since $BX$ is midpoint of $AB$, $AB = CD = 2$ . Now $AD$ and $BC$ remain to be calculated. How can the right angle $CXD$ be used to do so?

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Let $Y$ be the mid-point of $CD$. By symmetry, $\angle BYA$ = $90^\circ$. So $BYA$ is a semicircular arc. $XB$, $XY$, and $XA$ are all radii of this semicircle, with length $1$. But $XY$ is the average of the lengths $BC$ and $AD$. So $BC + AD = 2$.

Edited to add: In case anybody was wondering, there are many such trapezia. To see this, start from the OP's diagram (with $\angle CXD < 90^\circ$), and shorten $BC$ and $AD$ while keeping the slopes of $AB$ and $CD$ unchanged, until $C$ coincides with $B$. Now $\angle CXD > 90^\circ$, so somewhere in between, we must have had $\angle CXD = 90^\circ$. This works whenever $60^\circ < \angle BAD \le 90 ^\circ$.

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I verified your answer i.stack.imgur.com/ATMmD.png – schooler Dec 6 '11 at 11:03
@schooler: But your diagram is not symmetric! So my argument doesn't apply to it. – TonyK Dec 6 '11 at 11:12

If this is for a quiz, i.e. you know there is a single solution, then you can take an example of a $1\times 2$ rectangle which obviously fits the conditions and has a perimeter of $6$.

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Hey you got the right answer! but can you elaborate more? – schooler Dec 6 '11 at 10:05
+1 for deviousness – user7530 Dec 6 '11 at 11:04
@schooler: you are not given enough information to construct the trapezium, as there is more than one possibility with the conditions given. But if you think that the perimeter is invarient then you can choose any example to calculate its perimeter, and I chose an easy one. – Henry Dec 7 '11 at 0:49
I only gave info that was in the question – schooler Dec 7 '11 at 11:43