# [Euclidean Geometry] Kiselev's Plainimetry Question 242

In summary, the conversation discusses a geometry problem involving tangents and a circle. It is stated that the perimeter of triangle DME and the angle DOE do not depend on the position of point C. Two solutions are presented, one using the concept of bisectors and the other using the properties of kites. It is mentioned that these solutions can be used as a method for similar problems in the future.

## 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?

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.

RUber said:
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.

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.

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.

RUber said:
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.

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.

## 1. What is Kiselev's Plainimetry Question 242?

Kiselev's Plainimetry Question 242 is a famous problem in Euclidean Geometry that asks for the ratio of the areas of two triangles formed by the intersection of the diagonals of a quadrilateral.

## 2. How do you solve Kiselev's Plainimetry Question 242?

The solution to Kiselev's Plainimetry Question 242 involves using the properties of similar triangles, as well as the formula for the area of a triangle (A = 1/2 * base * height). It can also be solved using algebraic equations and the Pythagorean theorem.

## 3. What are the applications of Kiselev's Plainimetry Question 242?

Kiselev's Plainimetry Question 242 is a classic problem that helps students develop their understanding of geometric concepts and problem-solving skills. It also has real-world applications in fields such as architecture, engineering, and cartography.

## 4. Are there any variations of Kiselev's Plainimetry Question 242?

Yes, there are many variations of Kiselev's Plainimetry Question 242 that involve different types of quadrilaterals or ask for different relationships between the areas of the triangles. These variations can increase in difficulty and provide further practice in applying geometric principles.

## 5. Can Kiselev's Plainimetry Question 242 be solved using other methods?

Yes, there are multiple methods for solving Kiselev's Plainimetry Question 242, such as using trigonometric ratios, coordinate geometry, and even geometric constructions. However, the most common and efficient method involves using the properties of similar triangles.

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