пусть m – точка пересечения диагоналей ac и bd четырёхугольника abcd. применим неравенство треугольника к треугольникам abc, adc, bad и bcd: ac < ab + bc, ac < da + dc, bd < ab + ad, bd < cb + cd. сложив эти четыре неравенства, получим: 2(ac + bd) < 2(ab + bc + cd + ad).
запишем неравенства треугольника для треугольников amb, bmc, cmd и amd: am + mb > ab, bm + mc > bc, mc + md > cd, ma + md > ad. сложив эти неравенства, получим: 2(ac + bd) > ab + bc + cd + ad.
Speaking Task 1. Choose the question from the card on the topic Entertainment and fedia^ prime prime and be ready to answer it after the teacher starts the conversation. Produce a speech by giving extended answers to the questions. Share your ideas with the class. Teacher organizes a Socratic seminar, which helps him/her to assess learners while they are speaking on the toplic Entertainment and Media and he/she prepares and cuts down questions and expressions beforehand. Learners sit in a circle and answer the question using in their speech some formal and Informal expressions to present logically connected information to their classmates. Expressions: Stating an opinion The way I see it... Sorry to interrupt, but... Is it okay if I jump in for a second? Can I add something here? Can I throw my two cents in? Not necessarily Interrupting If I might add something..... I beg to differ No, I'm not so sure about that That's for sure Expressing disagreement I'd say the exact opposite I have to side with somebody (name)on this one I was just going to say that In my opinion Expressing agreement If you ask me.. That's exactly how I feel As far as I'm concerned. If you want my honest opinion..... You have a point there That's not always the case
пусть m – точка пересечения диагоналей ac и bd четырёхугольника abcd. применим неравенство треугольника к треугольникам abc, adc, bad и bcd: ac < ab + bc, ac < da + dc, bd < ab + ad, bd < cb + cd. сложив эти четыре неравенства, получим: 2(ac + bd) < 2(ab + bc + cd + ad).
запишем неравенства треугольника для треугольников amb, bmc, cmd и amd: am + mb > ab, bm + mc > bc, mc + md > cd, ma + md > ad. сложив эти неравенства, получим: 2(ac + bd) > ab + bc + cd + ad.