Cho tam giác ABC nhọn. Dựng phía ngoài tam giác ABC các hình vuông ABGF, ACDE, BCRQ. Gọi O1 , O2 , O3 theo thứ tự là tâm các hình vuông bên. a) Chứng minh: O1O2 vuông góc với AO3
b) Chứng minh: O1C; O2B; O3A đồng quy
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2 tháng 8 2019
hình = link
Gọi O1, O2, O3 lần lượt là tâm các hình vuông dựng từ cách cạnh AB, AC, BC
Ta thấy \(\Delta ACP=\Delta MCB\)(c-g-c) do AC=MC, gócACP=gócMCB, CP=BC => AP = BM
Gọi I, H lần lượt là giao điểm của BM với AP, AC
Xét 2 tam giác AIH và MCH có: góc AHI=góc MHC(đối đỉnh), góc IAH=góc CMH (do gócPAC=gócBMC)
=> \(\Delta AIH~\Delta MCH\) => \(\widehat{AIH}=\widehat{MCH}=90^0\) => AP vuông góc BM
Gọi D là trung điểm của AB, ta có:
Tam giác ABP có: DA=DB, O3B=O3P => DO3 là đường trung bình => DO3//AP và DO3=AP/2 (1)
Tam giác BAM có: DA=DB, O2A=O2M => DO2 là đường trung bình => DO2//BM và DO2=BM/2 (2)
(1) và (2) suy ra: DO3 vuông góc DO2 và DO3=DO2 (do AP vuông góc BM và AP=BM)
Dễ dàng thấy: tam giác DO1A = tam giác DO1B(c-c-c) => \(\widehat{ADO_1}=\widehat{BDO_1}=\frac{180^0}{2}=90^0\)
Có: \(\widehat{ADO_1}=\widehat{O_2DO_3}\)\(\left(=90^0\right)\)
\(\Leftrightarrow\)\(\widehat{ADO_1}+\widehat{ADO_2}=\widehat{O_2DO_3}+\widehat{ADO_2}\)
\(\Leftrightarrow\)\(\widehat{O_1DO_2}=\widehat{ADO_3}\)
Tam giác vuông DO1A có góc \(\widehat{AO_1D}=180^0-\left(\widehat{ADO_1}+\widehat{DAO_1}\right)=180^0-\left(90^0+45^0\right)=45^0\)
=> tam giác DO1A vuông cân tại D => DO1=DA
Xét 2 tam giác O1DO2 và ADO3 có: góc O1DO2 = góc ADO3(CM trên), DO1=DA(CM trên), DO2=DO3(đã CM ở đầu bài)
=> \(\Delta O_1DO_2=\Delta ADO_3\left(c-g-c\right)\) => \(O_1O_2=AO_3\)
CH
Cô Hoàng Huyền
Admin
VIP
21 tháng 11 2016
a. Ta thấy \(\widehat{EAC}=\widehat{BAH}\left(=\widehat{BAC}+90^o\right)\)
Vậy nên \(\Delta EAC=\Delta BAH\left(c-g-c\right)\)
Từ đó suy ra \(\widehat{ACE}=\widehat{AHB}\)
Vì \(\widehat{AHB}+\widehat{JHF}+\widehat{F}+\widehat{FCA}=270^o\Rightarrow\widehat{ACE}+\widehat{JHF}+\widehat{F}+\widehat{FCA}=270^o\Rightarrow\widehat{HJC}=90^o\)
Vậy \(EC\perp BH.\)
b. Ta thấy \(O_1\) là trung điểm EB. Vậy thì O1I là đường trung bình của tam giác BEC hay O1I // EC. Tương tự O2I // BH.
Lại có \(EC\perp BH\) nên \(O_1I\perp O_2I.\)
Vậy tam giác O1O2I là tam giác vuông tại I.