Tâm đường tròn ngoại tiếp là giao điểm của ba đường trung trực. Đúng hay sai?

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Theo đề bài ta có đường tròn ngoại tiếp ΔABC Δ A B C là đường tròn tâm O(0;0) O ( 0 ; 0 ) và bán kính R=√50=5√2. hay (ˆH1+ˆH2)+(ˆA1+ˆA2)=180° H 1 ^ + H 2 ^ + A 1 ^ + A 2 ^ = 180 ° , tức là ˆBHC+ˆJAK=180° B H C ^ + J A K ^ = 180 ° . Mar 4, 2025 ... Join this channel to get access to perks: https://www.youtube.com/channel/UCFhqELShDKKPv0JRCDQgFoQ/join Trong không gian Oxyz,cho tam giác ... Dec 20, 2024 ... ... cho tam giác ABC có điểm A(1;2;3),B(2;-1;3) và C(-1;1;1) Biết rằng tọa độ của chân đường cao hạ từ A xuống BC là H(a;b;c),giá trị của P=17(a+b+c) Nov 28, 2020 ... cho tam giác ABC có AB

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