0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ it can be shown that $\angle DOE=90^\circ+\frac{\angle BMA}2$ and the perimeter of triangle $DME$ is $2 \times MA$ $\endgroup$
    – timon92
    Commented Sep 2, 2015 at 18:56

1 Answer 1

Reset to default
1
$\begingroup$

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)

enter image description here

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.

$\endgroup$
4
  • $\begingroup$ How did you draw that diagram? $\endgroup$
    – fgnxfnxfgnxn
    Commented Sep 3, 2015 at 9:42
  • $\begingroup$ Using Geogebra. Image was then sent to WORD via the clipboard. After re-sizing, the adjusted image is then sent to PC paintbrush for the final touch. It is then uploaded. $\endgroup$
    – Mick
    Commented Sep 3, 2015 at 11:06
  • $\begingroup$ Thanks for the answer. I would upvote it if I could. For anyone that goes through this answer, I'd like to add that viewing $AODC$ and $BOCE$ as kites also helps in seeing the answer. $\endgroup$
    – fgnxfnxfgnxn
    Commented Sep 3, 2015 at 18:29
  • $\begingroup$ @modularnix More importantly, $AODC$ and $BOCE$ are cyclic quadrilaterals. Such quadrilateral processes more properties than a kite. $\endgroup$
    – Mick
    Commented Sep 4, 2015 at 3:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .