Math Help - Circle Geometry

1. Circle Geometry

This last question on my homework is really annoying. I'm pretty sure the answer is laughing right in my face, but I just can't do it.

The question is:

Ab and CD are two intersecting chords of a circle and CD is parallel to the tangent to the circle at B

I got a feeling that I need to prove that ABCD is a square/rhombus?

2. I saw this a while ago before I did something important. I knew right away that AB should be a diameter, etc, but I coudn't recall a rule/law/theorem/whatever that will say AB is a perpendicular bisector of chord CD.

Well, I got it now (wonder of internet): it is a "conjecture" that says, Conjecture (Perpendicular Bisector of a Chord ): The perpendicular bisector of a chord in a circle passes through the center of the circle.

Okay. Oh, I know that a radius that is perpendicular to a chord bisects that chord. But in Geometry, you have to say the reason why if somebody is concern or asking. Cannot bluff in Geometry. That's why I hate proofs.

line segment AB = line segment AB.
In Geometry, prove that, or say why is that.

Anyway, if B is the point of tangency, then AB is perpendicular to that external tangent line at B. So AB passes through the center and AB is a diameter too. At the point of tangency, one and only one chord is perpendicular to to tangent line. This chord must pass through the center of the circle.

Now if CD is parallel to the said tangent line, then AB must be perpendicular also to CD.

Then, the "Conjecture".
So AB is the perpendicular bisector of CD.
So in triangle CAD, side CD is equally divided into two.
So the base angles are equal.
So, by sum of all interior angles of a triangle is 180 degrees, angle CAB = angle DAB.

3. Hello, noreaction!

I'm pretty sure the answer is laughing right in my face. . I know the feeling!

$AB$ and $CD$ are two intersecting chords of a circle
and $CD$ is parallel to the tangent to the circle at $B$

Prove that $AB$ bisects $\angle CAD$.
It's much easier than you think . . .

$CD$ is parallel to the tangent at $B$.
Parallel chords intercept equal arcs (even if one chord is a tangent).
Hence: . $\text{arc}(BC) = \text{arc}(BD)$

$\angle BAC$ is an inscribed angle measured by $\frac{1}{2}\text{arc}(BC)$

$\angle BAD$ is an inscribed angle measured by $\frac{1}{2}\text{arc}(BD)$

Got it?