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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)
=> \(\Delta O_1DO_2=\Delta ADO_3\left(c-g-c\right)\) => \(O_1O_2=AO_3\)
Tứ giác EGCD có :
góc EBC = góc GCB = góc EGC = 90 độ
-> EGCB là hình chữ nhật
Mà P,Q,M,N lần lượt là đỉnh của 4 cạnh
->MNPQ là hình vuông
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Gọi M là trung điểm BC ; N là điểm đối xứng với H qua M.
M là trung điểm của BC và HN nên BNCH là hình bình hành
\(\Rightarrow NC//BH\)
Mà \(BH\perp AC\Rightarrow NC\perp AC\)hay AN là đường kính của đường tròn ( O )
Dễ thấy OM là đường trung bình \(\Delta AHN\) suy ra \(OM=\frac{1}{2}AH\)
M là trung điểm BC nên OM \(\perp\)BC
Xét \(\Delta AHG\)và \(\Delta OGM\)có :
\(\widehat{HAG}=\widehat{GMO}\); \(\frac{GM}{GA}=\frac{OM}{HA}=\frac{1}{2}\)
\(\Rightarrow\Delta AGH~\Delta MOG\left(c.g.c\right)\Rightarrow\widehat{AGH}=\widehat{MGO}\)hay H,G,O thẳng hàng
gọi E,F,T lần lượt là trung điểm của AB,CD,BD
Đường thẳng ME cắt NF tại S
Vì AC = BD \(\Rightarrow EQFP\)là hình thoi \(\Rightarrow EF\perp PQ\)( 1 )
Xét \(\Delta TPQ\)và \(\Delta SEF\)có : \(ME\perp AB,TP//AB\)
Tương tự , \(NF\perp CD;\)\(TQ//CD\)
\(\Rightarrow\Delta TPQ~\Delta SEF\)( Góc có cạnh tương ứng vuông góc )
\(\Rightarrow\frac{SE}{SF}=\frac{TP}{TQ}=\frac{AB}{CD}\)
Mặt khác : \(\Delta MAB~\Delta NCD\Rightarrow\frac{AB}{CD}=\frac{ME}{NF}\)( tỉ số đường cao = tỉ số đồng dạng )
Suy ra : \(\frac{ME}{NF}=\frac{SE}{SF}\)\(\Rightarrow EF//MN\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra \(MN\perp PQ\)
tự kẻ hình :
có M; N lần lượt là trung điểm của AB; AC (gt)
=> MN là đường tb của tam giác ABC (đn)
=> MN // BC (đl)
góc BCNM là tứ giác
=> BCNM là hình thang (đn)
a) Phép quay tâm C góc 90 ο biến MB thành AI. Do đó MB bằng và vuông góc với AI. DP song song và bằng nửa BM, DO song song và bằng nửa AI. Từ đó suy ra DP bằng và vuông góc với DO.
b) Từ câu a) suy ra phép quay tâm D, góc 90 ο biến O thành P, biến A thành Q. Do đó OA bằng và vuông góc với PQ.