I need help with a high school geometry proof. I think I've figured out why the prompt is true, but the proof attempt I've come up with seems very inelegant. Is there an easier method I'm missing?
Consider two circles with the second internally tangent to the first at point $A$ and also passing through the center of the first. Show that every chord of the first circle which has $A$ as an endpoint is bisected by the second circle.
My attempted proof:
Let there be any chord of the first circle which has $A$ as an endpoint. Let the other endpoint of the chord be called $B$.
Then let the following line segments be drawn:
- A segment connecting $A$ and the first circle's center $C$;
- A segment connecting $B$ and the first circle's center $C$; and
- A segment connecting the first circle's center $C$ with the point $I$ where the chord intersects the second circle.
Segments $AC$ and $BC$ are the same length because they both represent the radius of the first circle.
We have two right triangles $ACI$ and $BCI$. Since the two hypotenuses $AC$ and $BC$ are the same length and the two heights $CI$ are the same length, then the two bases $AI$ and $BI$ must also be the same length. Since $AI$ is half the length of the chord, the second circle bisects the chord.
Thanks in advance for your help.
Best Answer
Another proof: Since both CI and BD are perpendicular to AB, CI is parallel to BD. The fact that C is the midpoint of AD implies that I is the midpoint of AB.