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Show that, for any values of x and y where x>y,
the angle SPQ is always the same and find this angle
the angle SRQ is always the same and find this angle
the quadrilateral SPQR is cyclic
the angle PQR is always the same and find this angle
the angle PSR is always the same and find this angle.Source: Nrich.maths.org
SPQ is a right angle because triangle QBP is similar to triangle PAS by side-angle-side, so angle APS + angle BPQ is 90�, so angle SPQ is also 90� because the sum of all three angles is 180�. Another way to see this is by observing that SQ2=PS2+PQ2:
PS2 = AP2+AS2 = y2(x-y)2 + y2(x+y)2 = 2y2(x2+y2)
PQ2 = BP2+BQ2 = x2(x-y)2 + x2(x+y)2 = 2x2(x2+y2)
SQ2 = (AS-BQ)2+(AP+PB)2 = (y(x+y)-x(x-y))2+(y(x-y)+x(x+y))2 = 2(x2+y2)2 = PS2+PQ2
QRS is a right angle because just as in the previous case, triangle SDR is similar to triangle RCQ. Also, SQ2=RQ2+RS2:
RQ2 = CR2+CQ2 = x2(x+y)2+y2(x+y)2 = (x2+y2)(x+y)2
RS2 = DS2+DR2 = x2(x-y)2+y2(x-y)2 = (x2+y2)(x-y)2
RQ2+RS2 = (x2+y2)((x+y)2+(x-y)2) = 2(x2+y2)2 = SQ2
The quadrilateral SPQR is cyclic because there is only one circle with diameter SQ, and this is the circumcircle both of triangle SPQ and of triangle SRQ. As this circle contains all four points S,P,Q,R, it follows that SPQR is cyclic.
For convenience, I will refer to the circumcircle of quadrilateral SPQR as "circle O". As circle O's diameter is SQ, and SD=QB, it is clear that the center of circle O, which I will call point O, is also the center of the rectangle ABCD, i.e. the intersection of its diagonals.
To see that PQR is a 45� angle,
Circle O intersects the long sides of rectangle ABCD at four points. These points are R, P, and their reflections about point O, which I will call R' and P'. By symmetry, R' is on line segment PB, at a distance of y(x-y) from B. P' is on line segment RC, at a distance of y(x-y) from C. Now observe that RPR'P' is a square whose sides are x2+y2 units long. This square is inscribed in circle O, cutting it into four equal arcs. Angle PQR is inscribed in circle O, and is subtended by one quarter of this circle.
Last, angle PSR is the supplement of angle PQR, so it is 135�.
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