Look at the image uploaded below. Because `BC = CD , ` the triangle `DCB ` is isosceles and `CP ` is the height, not only the median of `DCB .`

Also, the angles `CDB ` and `DBA ` are equal as alternate interior ones. The lines `AD ` and CP are perpendicular to the line `BD , ` so they are parallel. Because of this, the angles `PCA ` and `DAC ` are also alternate interior and therefore equal.

Denote `CP = h , ` `AD = x , ` `DP = BP = y . ` We can construct three equations:

`tan ( PCA ) = 11 / h = ( y - 11 ) / x ,`

`tan ( ADB ) = x / ( 2y ) = h / y ,`

and `h^2 + y^2 = 43^2 .`

It is easy to solve: `h = x / 2 , ` `22 / x = ( y - 11 ) / x , ` so `y = 33 . ` From the first equation `( x / 2 )^2 = 43^2 - 33^2 ` and

`x = 2 sqrt ( 43^2 - 33^2 ) = 2 sqrt ( ( 43 - 33 ) ( 43 + 33 ) ) = 2 sqrt ( 10 * 76 ) = 4 sqrt ( 190 ) .`

Now we see that `m = 4 ` and `n = 190 , ` so the answer is **194**.

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