Two lines passing through a point $Μ$ are tangent to a circle at the points $A$ and $B$. Through a point $С$ taken on the smaller of the arcs $AB$, a third tangent is drawn up to its intersection points $D$ and $Ε$ with $MA$ and $MB$ respectively. Prove that (1) the perimeter of triangle $DME$, and (2) the angle $DOE$ (where $О$ is the centre of the circle) do not depend on the position of the point $C$.
(This is question 242 from Kiselev's Geometry, Book I: Planimetry.)
I think that the angle $DOE$ will be equivalent to the angle $AOM$ or angle $BOM$ (both of them are equal). I've reasoned this by imagining the point $C$ sliding on the smaller arc $AB$ and when $C$ is at point $A$ and $B$, angle $DOE$ becomes equivalent to angle $AOM$ or $BOM$.
I've come up with a (probably) flawed reason for proving the first part of the question. Again, I'm imagining $C$ sliding on the smaller arc $AB$ like a pendulum would and because triangle $AOM$ is congruent to the triangle $BOM$, the triangle $DME$ would swipe out equal areas in the triangles $AOM$ and $BOM$.
I've probably used faulty reasoning and I'm not convinced that I've solved this question. Would someone like to give me a hint as to how I should proceed?
This is exactly the comment made by @timon92. I supplement it with a diagram with short explanation. Hope this can be of further help.
Preliminaries:- (1) By tangent properties, $DA = DC$, $EC = EB$, $MA = MB$; (2) $\beta_1 = \beta_2$, $\gamma_1 = \gamma_2$; (3) $\alpha = \angle M$ (exterior angle of cyclic quadrilateral)
Perimeter of $\triangle MDE = (ME + CE) + (MD + DC)$
$ = (ME + EB) + (MD + DA)$
$ = 2MA$
$\beta_2 + \gamma_1 = \dfrac {2\beta_2 +2\gamma_1}{2} = \dfrac {180^0 – \alpha}{2} = \dfrac {180^0 – \angle M}{2}$, a constant.