Help Please I am stuck

SloMo150

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Joined
Apr 3, 2006
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Well I am working on a doozy....

I am stuck after a point


Original problem:

2 congruent circles tangent at X. Centers A and B Respecively

Chord - XC (Circle A) is perpendicular to Chord XD (Circle B)

Prove AB is parallel to CD and that AB = CD



OK I have that C is complimentary to D but I am getting stuck... you all are so quick I hope that you can help me finish this up nice and neat
 
I have that C is complimentary to D Let's add two more points, E & F where AB meets the rims of A & B respectively. That makes < ECX and XDF right angles (as triangles in circles with the diameter as hypots are) and makes EC parallel to XD and XC parallel to DF. We then have three congruant triangles (ASA)
Can you finish it off from there?
 
Hello, SloMo150!

I did it by measuring angles . . .

Two congruent circles tangent at X; centers A and B, respecively.
Chord XC (circle A) is perpendicular to chord XD (circle B).
Prove: ABCD\displaystyle AB\,\parallel\,CD and AB=CD\displaystyle AB\,=\,CD.
Code:
                  C                   D
              * * o               * * o
          *         \ *       *     /     *
        *            \  *   *     /         *
       *              \  * *    /            *
                       \      /
      *                 \ * /                 * 
      *         o - - - - o - - - - o - - - - o E
      *         A         *X        B         *

       *                 * *                 *
        *               *   *               *
          *           *       *           *
              * * *               * * *
We are given: CXD=90o\displaystyle \,\angle CXD\,=\,90^o

Draw radii AC=BD=r\displaystyle AC\,=\,BD\,=\,r

Extend AB\displaystyle AB\, to E\displaystyle E.

Let CAX=2θ\displaystyle \angle CAX\,=\,2\theta
Since ΔCAX\displaystyle \Delta CAX is isosceles, CXA=90oθ\displaystyle \angle CXA = 90^o\,-\,\theta

Since AXB=180o\displaystyle \angle AXB\,=\,180^o, we have: DXB+90o+(90oθ)=180o        DXB=θ\displaystyle \,\angle DXB\,+\,90^o\,+\,(90^o-\theta)\:=\:180^o\;\;\Rightarrow\;\;\angle DXB\,=\,\theta

Since ΔXBD\displaystyle \Delta XBD is isosceles, XDB=θ        DBX=180o2θ\displaystyle \angle XDB\,=\,\theta\;\;\Rightarrow\;\;\angle DBX\,=\,180^o\,-\,2\theta

Then: DBE=2θ\displaystyle \,\angle DBE\,=\,2\theta

Hence: ACBD    \displaystyle \,AC\,\parallel\,BD\;\; (alternate-interior angles)


Therefore:

Quadrilateral ABDC is a parallelogram
    \displaystyle \;\;(if two sides of a quadrilateral are equal and parallel, the quadrilateral is a parallelogram)

and: ABCD\displaystyle AB\,\parallel\,CD and AB=CD\displaystyle AB\,=\,CD
    \displaystyle \;\;(opposite sides of a parallelogram are parallel and equal).
 
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