Geometry Intersecting Circles and Secants

[skinski43]

I'm having problems solving this proof. Two circles intersect at the points A and B, and through A, a secant is drawn intersecting the circles at the points C and D. Prove that the measure of the angle CBD is constant, i.e. it is the same for all such secants. When I draw it out on MS Publisher it looks like the measure of the angle CBD is truly the same. I measure the angles on the monitor, but I don't really understand why.

Bernie

 

[soroban]

Hello, sBernie!

Two circles intersect at the points A and B.
Through A, a secant is drawn intersecting the circles at the points C and D.
Prove that the measure of the angle CBD is constant.

              * * *   A   * * *
        C o - - - - - o - - - - - o D
        *  \        *   *        /  *
       *    \      *     *      /    *
             \                 /
      *        \  *       *  /        *
      *          \*       */          *
      *         P o       o Q         *
                   \     /
       *           *\   /*           *
        *           *\ /*           *
          *           o           *
              * * *   B   * *

\(\displaystyle \text{The two arcs from }A\text{ to }B\text{ are constant.}\)

\(\displaystyle \angle C \:=\:\tfrac{1}{2}\widehat{AQB},\:\text{ constant}\)

\(\displaystyle \angle D \:=\:\tfrac{1}{2}\widehat{APB},\:\text{ constant}\)

\(\displaystyle \text{In }\Delta CBD\!:\;\;\angle CBD + \angle C + \angle D\:=\:180^o\)

\(\displaystyle \text{Therefore: }\:\angle CBD \;=\;180^o - \angle C - \angle D \;=\;\text{constant}\)