Geometric Proof Involving Circles скачать в хорошем качестве

Geometric Proof Involving Circles

#geometry #proofs #WACE The solution begins at 3:00 I hope it helps :) A worked solution is available here: https://drive.google.com/file/d/1R3Yy... When a secant (meeting the circle at A and B) and a tangent (meeting the circle at T) are drawn to a circle from an external point M, the square of the length of the tangent equals the product of the lengths to the circle on the secant (AM × BM = TM^2). With these proofs, those who struggle will often look at people that these kinds of problems appear to come easy to; whether it be their classmates, tutors, or even their teachers. In my experience, when any of these people are asked how they do it, their typical response is one of two things; 1) I don’t know man, I just muck around with the rules that I know until something comes of it, or 2) I kind of just see the problem’s solution from the get-go. So, unfortunately, the first option is not ideal under timed conditions such as in a test or exam, I think that there is a deliberate method that you can go about as a means to prompt the solution – think of it as giving yourself a hint. So, what I’ll do is try and break down the things that I now intuitively do to try and solve a problem of this nature. So, because the TM is squared and part of the triangle MTC, for me that is a good indication that the Pythagorean theorem should be used as part of the proof – so let’s start with that. Now the point B on the diagram is not at all part of the triangle MTC however, the way we can incorporate that into our situation is by using the fact that in this triangle the hypotenuse can be split up into two parts with one of them being the radius of the circle.