Prove that the points D, B, X, and Y are concyclic

### 1 Answer

Let for convenience |AB| = |CD| =a, |AD| = |BC| = b, |BX| = x, |DY| = y, then:

(1) a/y = b/x (triangle DCY is similar to BXC);

(2) a^2 = by (CD is a height in the right triangle ACY);

(3) b^2 = ax (BC is a height in the right triangle ACX).

The statement is equivalent to the following: the angles XBY and XDY to be equal (XY to be seen from B and D at the same angle), the latter equivalent to: complement angles ADX and ABY to be equal but we have:

tan(ABY) = (b + y)/a, but y = a^2/b according (2);

tan(ADX) =(a + x)/b, but x = b^2/a according (3), then

(b + y)/a = (b + a^2/b)/a = (b^2 + a^2)/(ab)

(a + x)/b = (a + b^2/a)/b = (a^2 + b^2)/(ab), what completes the proof.

P.S.(Edit – Analytic geometry – second proof) In above notation take A as origin of the coordinate system, X-axis along ABX, Y-axis along ADY, then:

A(0, 0), B(a, 0), C(a, b), D(0, b),

X(a + b^2/a, 0), Y(0, b +a^2/b), take the point

Z(a + b^2/(2a), b + a^2/(2b)), then the check easily produces

|ZX| = |ZY| = |ZB| = |ZD|, what means that Z is a center of a required circle.