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Explanation As a real droodle, the configuration presented in the applet admits several interpretations. For example, let Q, R be two points on a diameter CP of a circle with center O. Draw chords KN through Q and ST through R perpendicular to PC. Let SQ intersect the circle at V. Finally, let X denote the intersection of PV and CT. Then X lies on KN.

Here is another interpretation. Given a circle with diameter CP and center O. Points V and T are selected on one side of CT and PV cross at X. QX, and RT are perpendicular to CP. S is the point of intersection of RT with the circle. Then SV passes through Q.

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Let's address the fist problem. Join CV and PT. Since angles CVP and CTP are subtended by the diameter CP, they are right. Thus in ΔCPX the chord CV and PT are altitudes. The line through their point of intersection and X is then the third altitude. (We encountered a similar circumstance in a Sangaku problem.)

Assume, S is a reflection of T in PC and V is the intersection of SQ and the circle. Let TQ meet the circle at U. Angles at S and T are equal by the construction and so are the arcs SU and TV they subtend. ∠TQV is half the sum of the angular measures of these arcs. Therefore, ∠TQV = 2∠TSV say. Further, KN||ST implies

 ∠NQV = ∠TSV (as corresponding angles) and ∠NQT = ∠STU (as alternate interior angles)

With these, we may conclude that QN bisects ∠TQV in ΔTQV. And because T and V are the feet of two altitudes in ΔCPX, so is Q, for no other point on the base CP could form an angle with T and V that is bisected by the perpendicular to the base. Thus XQ is the third altitude in ΔCPX and since QN is a part of that altitude, X lies on KN. 