Hello I've bean practicing for competition in math and can't seem to solve this problem,tried drawing chords,tangents,finding equal triangles,but couldn't seem to solve it.Any help would be appreciated.
Circles $k_1$, with center in $O_1$ and radius $r$ and $k_2$, with center in $O_2$ and radius $2r$, touch internally.Chord $AB$ of circle $k_2$ touch circle $k_1$ in point $T$.Let $p$ be a normal line from $O_2$ onto $AB$, and let her second intersection with circle $k_1$ be point $C$. Let $D$ be the point of intersection of $p$ and $k_2$ which is on the opposite side of $O_2$ in relation to $AB$.Prove that the line $AB$ is a perpendicular bisector of $CD$
$\begingroup$
$\endgroup$
7
-
1$\begingroup$ A drawing here would certainly help a lot... $\endgroup$DonAntonio– DonAntonio2014-02-23 16:04:55 +00:00Commented Feb 23, 2014 at 16:04
-
$\begingroup$ I'll add the drawing as soon as I get to my computer. $\endgroup$kingW3– kingW32014-02-23 16:10:38 +00:00Commented Feb 23, 2014 at 16:10
-
$\begingroup$ Added the picture,hope it's clearer now $\endgroup$kingW3– kingW32014-02-23 16:25:41 +00:00Commented Feb 23, 2014 at 16:25
-
$\begingroup$ Good drawing, though slightly too small. The problem is I've no idea what "a line symmetry" of a cord, or of a line, is...never heard of it within this context. $\endgroup$DonAntonio– DonAntonio2014-02-23 16:31:02 +00:00Commented Feb 23, 2014 at 16:31
-
1$\begingroup$ So "line of symmetry" in this case means "perpendicular bisector" $\endgroup$abiessu– abiessu2014-02-23 16:52:31 +00:00Commented Feb 23, 2014 at 16:52
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
Call the point where $AB$ and $CD$ intersect $E$ and the point where the circles touch $F$. Then the line $FT$ also goes through $D$, as the lines defined by $TO_1$ and $DO_2$ are both mutually perpendicular to the line $AB$ and so are parallel. Note that $T$ lies on the circle for which $O_2F$ is the diameter, so angle $O_2 T F$ is a right angle, and so triangle $O_2 T D$ is a right triangle.
By similarity, that means that $\frac{EO_2}{ET} = \frac{ET}{DE}$, or $DE\cdot EO_2 = ET^2$. Also, by the secant-tangent theorem, $EO_2 \cdot EC = ET^2$, so $EC = ED$.
