# 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?

-

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 $60^\circ < \angle BAD \le 90 ^\circ$.

-
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 $6$.