The Book of Lemmas: Proposition 9
If in a circle two chords AB, CD intersect at right angles, then:
(arc AD) + (arc CB) = (arc AC) + (arc DB).
Adjust the sliders to change the horizontal and vertical positions of DC and AB, xDC and yAB respectively. Observe that the sum of the lengths of arcs AD and CB and arcs AC and DB are always equal to half of the circle's circumference, π, where the circle's radius = 1.
(arc AD) =
(arc CB) =
(arc AC) =
(arc DB) =
r (arc AD) + (arc CB) = (arc AC) + (arc DB)
Let the chords intersect at O, and draw the diameter EF parallel to AB intersecting CD in H. EF will thus bisect CD at right angles in H, and:
(arc ED) = (arc EC).
Also EDF, ECF are semicircles, while:
(arc ED) = (arc EA) + (arc AD).
(sum of arcs CF, EA, AD) = (arc of a semicircle).
And the arcs AE, BF are equal. Therefore:
(arc CB) + (arc AD) = (arc of a semicircle).
Hence the remainder of the circumference, the sum of arcs AC, DB is also equal to a semicircle; and the proposition is proved.
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