Cho tam giác ABC có 3 góc nhọn , AB<AC , Ai là đường phân giác trong của tam giác ABC , I thuộc BC . Đường thẳng vuông góc với AI tại I cắt tia đối của tia BA tại M và cắt AC tại N . So sánh MN và BC
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Gọi ( O;R ) , ( I ;r ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF};\widehat{BAC}=\widehat{EDF}\)) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ACB},\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)( hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OB\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
\(ID=IE\left(=r\right)\Rightarrow\Delta IDE\)cân tại I
Do đó Tam giác OAB ~ Tam giác IDE \(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\) ( đpcm)
Gọi ( O; R ), ( I; R ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF;}\widehat{BAC}=\widehat{EDF}\) ) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ABC}=\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)(hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OA\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
Do đó Tam giác OAB ~ Tam giác IDE\(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\left(đpcm\right)\)
Rất vui vì giúp đc bạn <3
Gọi AJ là đường trung tuyến của \(\Delta\)ABC. Đường thẳng qua N song song AB cắt BC tại P.
Đường thẳng qua C song song AB cắt đường thẳng qua M song song BC và AJ lần lượt tại Q,R.
Ta thấy \(\Delta\)MAN có đường cao AI đồng thời là đường phân giác nên \(\Delta\)MAN cân tại A
=> I cũng là trung điểm cạnh MN. Từ đó \(\Delta\)MBI = \(\Delta\)NPI (g.c.g) => NP = BM; ^INP = ^IMB
Mà NP // BM // CQ, BM = CQ nên NP // QC, NP = QC => Tứ giác NPQC là hình bình hành
Nếu ta gọi K là trung điểm PC thì N,K,Q thẳng hàng
Chú ý rằng \(\Delta\)NPC ~ \(\Delta\)ABC (g.g) với trung tuyến tương ứng NK,AJ => \(\Delta\)NPK ~ \(\Delta\)ABJ (c.g.c)
=> ^PNQ = ^PNK = ^BAJ. Kết hợp với ^INP = ^IMB (cmt) suy ra ^MNQ = ^INP + ^PNQ = ^BAJ + ^IMB (1)
Mặt khác: \(\Delta\)ABJ = \(\Delta\)RCJ (g.c.g) => AB = CR < AC => ^BAJ = ^CRJ > CAJ
Điều đó có nghĩa là ^BAJ > ^BAC/2 = ^BAI => ^BAJ + ^IMB > ^BAI + ^IMB = 900 (2)
Từ (1) và (2) suy ra ^MNQ > 900 => MQ là cạnh lớn nhất trong \(\Delta\)QMN => MN < MQ = BC
Vậy MN < BC.