P and Q are two points on the diameter AB of a semicircle. K(T) is the circle tangent to the semicircle and the perpendiculars to AB at P and Q. Show that the distance from K to AB is the geometric mean of the lengths of AP and BQ.
Source: Anubhav Mishra
Let the points of tangency of XP on K(T) be R and of YQ on K(T) be S. Since T, K, O lie one a line, and the diameters RS and AB are parallel, by similarity we can see that A, R, T lie on one line, and B, S, T lie on another line. Since ATB is subtended by diameter AB, ATB is a right angle, and so ARP = SBQ and RAP = BSQ. Therefore triangle APR ~ triangle SQB, and their corresponding sides are proportional to each other, i.e. AP/SQ = RP/BQ. But RP = KC = SQ, so AP/KC = KC/BQ, or KC^2 = AP*BQ. Hence KC is the geometric mean of AP and BQ.
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