0
$\begingroup$

Given is a circle $k$ with diameter $AB$. On it, we choose a point $M$, which is not coincident with $A$ or $B$. Let $k_1$ be the circle that has its center at $M$ and is tangent to the diameter $AB$. Prove that the tangents from point $A$ and the tangent from point $B$ to the circle $k_1$ are parallel.

Attempt: First, I labeled the center of the circle $k$ with $S$ and the point of tangency of the circle $k_1$ with $AB$ with $N$. If I could somehow see that the angle at $A$ is equal to $180$ degrees minus the angle at $B$, that would be it. Any ideas on what to observe or what to connect? Thank you in advance for your help.

$\endgroup$
5
  • $\begingroup$ Note that this is equivalent to proving that the segment $\overline{A’B’}$, where$A’$ and $B’$ are the points of tangency, is a diameter of $k_1$. $\endgroup$
    – Paul Tanenbaum
    Commented Jun 18 at 12:24
  • $\begingroup$ Why exactly :)? $\endgroup$
    – user1315539
    Commented Jun 18 at 12:28
  • $\begingroup$ They’re tangents to the same circle, so they are parallel if and only if… [fill in the blank]. $\endgroup$
    – Paul Tanenbaum
    Commented Jun 18 at 12:32
  • $\begingroup$ I don't know how to fill it. It seems logical to me that A'B' has to through the center of k'. $\endgroup$
    – user1315539
    Commented Jun 18 at 12:35
  • $\begingroup$ Yes, and segments whose endpoints lie on a circle and that pass through the circles center are called diameters of the circle. To prove that my restatement is equivalent to the original problem, rotate, translate, and scale your coordinate system so that $k_1$ is the unit circle and consider at which two points horizontal (and thus parallel) lines are tangent to it (the unit circle). $\endgroup$
    – Paul Tanenbaum
    Commented Jun 18 at 12:42

1 Answer 1

Reset to default
1
$\begingroup$

Draw $MC$, where $C$ is the tangent of $AB$ and $k1$. $\angle{BMA}=90$ because it is inscribed and subtends a diameter. Define $\angle{BMC}=\beta, \angle{CMA}=\alpha$. Since tangents to a circle from a point are equal, we have $BD=BC$ and congruent triangles $MDB,MCB$ with $\alpha$ at $B$, and likewise for $\beta$ at $A$. Thus the internal opposite angles of $DBA,BAE$ sum to $2 \alpha+2 \beta=180$ and $DB, EA$ are therefore parallel.

enter image description here

$\endgroup$
4
  • $\begingroup$ I only don't see, why $BMC$ equals $\beta$ (and $CMA$ equals $\alpha$)? $\endgroup$
    – user1315539
    Commented Jun 18 at 13:08
  • $\begingroup$ Nice explanations, nice figure... $\endgroup$
    – Jean Marie
    Commented Jun 18 at 13:15
  • 1
    $\begingroup$ @good12 - I define angle BMC to be beta. Then it shows up in the angle at A because it is the complement of alpha, at angle CAM. If this isn't obvious, try it for any right triangle, it's a very useful property that the altitude to the right angle cuts the triangle into 3 similar triangles, you use this property over and over in geometry. $\endgroup$
    – RobinSparrow
    Commented Jun 18 at 13:20
  • $\begingroup$ I understand everything now perfectly, thanks a lot! $\endgroup$
    – user1315539
    Commented Jun 18 at 13:44

You must log in to answer this question.