The Book of Lemmas Proposition 9 - Maple Help

The Book of Lemmas: Proposition 9

 Main Concept If in a circle two chords AB, CD intersect at right angles, then: (arc $\mathrm{AD}$) + (arc $\mathrm{CB}$) = (arc $\mathrm{AC}$) + (arc $\mathrm{DB}$).

Adjust the sliders to change the horizontal and vertical positions of $\mathrm{DC}$ and $\mathrm{AB}$, $\mathrm{xDC}$ and $\mathrm{yAB}$ respectively. Observe that the sum of the lengths of arcs $\mathrm{AD}$ and $\mathrm{CB}$ and arcs $\mathrm{AC}$ and $\mathrm{DB}$ are always equal to half of the circle's circumference, $\mathrm{π},$ where the circle's radius = 1.

 $\mathbit{y}\mathbit{AB}$:   Arc Lengths: (arc $\mathrm{AD}$)            =       (arc $\mathrm{CB}$)            =       ∑                      =         (arc $\mathrm{AC}$)            =       (arc $\mathrm{DB}$)            =       ∑                      =         r (arc $\mathrm{AD}$) +  (arc $\mathrm{CB}$) = (arc $\mathrm{AC}$) + (arc $\mathrm{DB}$)   Proof: 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). Therefore:          (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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