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 $CD$
[Math] Two circles touch internally
circlesgeometry
Best Answer
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 $EC = ED$.