Help Please I am stuck

[SloMo150]

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

 

[Gene]

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?

 

[soroban]

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: \(\displaystyle AB\,\parallel\,CD\) and \(\displaystyle AB\,=\,CD\).

                  C                   D
              * * o               * * o
          *         \ *       *     /     *
        *            \  *   *     /         *
       *              \  * *    /            *
                       \      /
      *                 \ * /                 * 
      *         o - - - - o - - - - o - - - - o E
      *         A         *X        B         *

       *                 * *                 *
        *               *   *               *
          *           *       *           *
              * * *               * * *

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

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

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

Let \(\displaystyle \angle CAX\,=\,2\theta\)
Since \(\displaystyle \Delta CAX\) is isosceles, \(\displaystyle \angle CXA = 90^o\,-\,\theta\)

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

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

Then: \(\displaystyle \,\angle DBE\,=\,2\theta\)

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