[Euclidean Geometry] Kiselev's Plainimetry Question 242

[modularmonads]

Homework Statement

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 ▲DME, and (2) the ∠DOE (where О is the center of the circle) do not depend on the position of the point C.

2. The attempt at a solution
In my first attempt, I imagined the point C sliding on the smaller arc AB like a pendulum and when C is at A or B, the ∠DOE will be half of the ∠AOB. Following the same imagination, because the tangent at C will swipe out equal areas in the ▲s AOM and BOM, the perimeter of the ▲DME will remain constant.

In my second attempt, I followed a hint and proved that ∠DOE is half of ∠AOB and the perimeter of ▲DME is equal to MA+MB. However, I've proved this only when the tangent at C is perpendicular to OM. Even if this proof is all that is required, how shall I prove that the perimeter of ▲DME and the ∠DOE are independent of the position of point C?

 

[RUber]

Look at the arc AOD and compare it to the arc DOC, since OA and OC are both radii, OD is the bisector of that arc...no matter where C is.
Using this, you should be able to conclude, in the general sense that angle DOE is half of angle AOB, regardless of the location of C. And similarly, perimeter of DME = MA+MB.

 

[modularmonads]

Look at the arc AOD and compare it to the arc DOC, since OA and OC are both radii, OD is the bisector of that arc...no matter where C is.
Using this, you should be able to conclude, in the general sense that angle DOE is half of angle AOB, regardless of the location of C. And similarly, perimeter of DME = MA+MB.

Thanks for replying!

Oh, I didn't think about that. If we consider D and E to be points from which tangents are drawn (A and C from D and B and C from E) the quadrilaterals OADC and OCEB will be kites. Therefore, OD will bisect AC and OE will bisect BC at 90 degrees. This will be true no matter where C is.
Similarly, the perimeter of DME has been proven to equal to MA+MB and (still) I can only reason that the ascent on one side will always be equal to the descent on the other side in every other case. I don't know if this is sufficient.

 

[RUber]

For the perimeter, you can see from the kites that DC = DA and CE=EB so when you add up the sides, you still get MA+MB.

 

[modularmonads]

For the perimeter, you can see from the kites that DC = DA and CE=EB so when you add up the sides, you still get MA+MB.

Ah. I see. I feel bad for not being able to solve it on my own.

Thanks for your help!

 

[RUber]

These problems give you a lot of options and you have to find the best road to get the solution. I had to look at it for about an hour before I saw the right path. Don't feel bad, just add it to your bag of tricks for next time.