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 $\mathrm{DB}$).

$$$\mathit{y}\mathit{AB}$:
  

$$${\mathit{x}}_{\mathit{DC}}\mathbf{\:}$

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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