1. ## Obvious problem with quadrilateral

A circle is inscribed ABCD with the diameter 8.

AB = 12
CD = 8

AB is parallel to CD

Now looking at the picture here, it should be obvious that BC cannot be smaller than AD seeing as DK should always equal CP.

What am I misunderstanding here?

2. ## Re: Obvious problem with quadrilateral

Originally Posted by Paze

A circle is inscribed ABCD with the diameter 8.

AB = 12
CD = 8

AB is parallel to CD

Now looking at the picture here, it should be obvious that BC cannot be smaller than AD seeing as DK should always equal CP.

What am I misunderstanding here?
Revise your figure to show that the center of the circle must be on the bisectors of the angles.

3. ## Re: Obvious problem with quadrilateral

Originally Posted by votan
Revise your figure to show that the center of the circle must be on the bisectors of the angles.
It seems mathematically incorrect to assume that if CD is 8 and the diameter is 8...Then the sides CP and DK ara not parallel...They should by all means be parallel, right?

4. ## Re: Obvious problem with quadrilateral

I get it now. I'm misunderstanding that the tangent points do not necessarily mean the end points of my diameter.