49th Mathematical Olympiad in Ukraine

Question

In the rectangular trapezoid ABCD the side CD is perpendicular to the bases. Circle with diameter AB intersects AD at A and P, the tangent to this circle at the point P intersects line CD at M. From the point M the line l is drawn, which is tangent to our circle at the point Q. Prove that the line BQ halves the segment CD. FIGURE_0

Accepted Answer

Let N be the point of intersection of BQ and CD, L be the point of intersection of BQ and AD and O be the midpoint of AB. Denote the radius of our circle by r. We have that (fig.22) NBC = NLD = 12( AQ - BP) = 12( AOQ - BOP) = 12(180^ - QOB - BOP) = 90^ - 12( QOB + BOP) = 90^ - 12 QOP = FIGURE_0 Fig.22 90^ - QOM = QMO NBC QOM = POM, by two angles, thus CNBC = OQOM CN = BCOMr = PDPMr. Since BAP = BPM, as they share on the same arc, we deduce that ABP = MPD ABP MPD, and so CN = PDPMr = rABBP = 12BP = 12CD, what was to be proved.