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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
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)
=> \(\Delta O_1DO_2=\Delta ADO_3\left(c-g-c\right)\) => \(O_1O_2=AO_3\)
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a) xét tam giác ABC có:
. D là trung điểm của AB (gt)
. E là tđ của BC (gt)
vậy: DE là đường trung bình của tam giác ABC
--> DE//AC VÀ DE=\(\frac{AC}{2}\)
--> ACED là hình thang ( tứ giác có 2 cạnh đói //)
mà góc BAC=900 (tam giác ABC vuông tại A)
--> ACED là hình thang vuông( hình thang có 1 góc vuông)
b) Ta có: F đối xứng với E qua D (gt)
--> D là trung điểm của EF
--> EF=2DE
Ta lại có: DE=\(\frac{AC}{2}\) (cmt)
--> AC=2DE
Xét tứ giác ACEF có:
. DE//AC ( cmt)
--> EF//AC (D ϵ EF)
. EF=AC ( cùng = 2DE )
Vậy: ACEF là hbh (tứ giác có 2 cạnh đối vừa //, vừa = nhau)
c) Ta có: E là tđ của BC (gt)
--> CE=\(\frac{BC}{2}\) (1)
Ta lại có: E là tđ của BC (gt)
--> AE là đường trung tuyến
--> AE=\(\frac{BC}{2}\)
Xét tứ giác AEBF có:
.D là tđ của AB (gt)
. D là tđ của EF (cmt)
Vậy: AEBF là hbh( tứ giác có 2 đường chéo cắt nhau tại tđ của mỗi đường)
Ta có: AE= BF ( cạnh đối hbh AEBF)
mà AE=\(\frac{BC}{2}\) (cmt)
--> BF=\(\frac{BC}{2}\) (cùng = AE) (2)
Từ(1) và (2)
--> CE=BF (cùng =\(\frac{BC}{2}\) )
- Cách chứng minh của mình hơi dài nha ^.^