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a) Ta có: \(\left\{{}\begin{matrix}\widehat{BAC}+\widehat{EAB}=\widehat{EAC}\\\widehat{BAC}+\widehat{HAC}=\widehat{BAH}\end{matrix}\right.\) (1)
ABDE là hình vuông
=> Góc EAB = 900 (2)
Và: AE = AB
ACGH là hình vuông
=> Góc HAC = 900 (3)
Và: AH = AC
Từ (1); (2) và (3)
=> Góc EAC = Góc BAH
Xét ΔAEC và ΔABH ta có
AC = AH (cmt)
Góc EAC = Góc BAH (cmt)
AE = AB (cmt)
=> ΔAEC = ΔABH (c - g - c)
=> EC = BH (2 cạnh tương ứng) (4)
b) Có I là tâm của hình vuông ABDE
=> I là trung điểm của đường chéo EB
Lại có: M là trung điểm của BC
=> IM là đường TB của tam giác EBC
\(\Rightarrow IM=\frac{1}{2}EC\left(5\right)\)
Có J là tâm của hình vuông ACGH
=> J là trung điểm của đường chéo HC
Lại cos: M là trung điểm của BC
=> JM là đường TB của tam giác HBC
\(\Rightarrow JM=\frac{1}{2}BH\left(6\right)\)
Từ (4); (5); (6)
=> JM = IM
=> Tam giác JIM cân tại M (*)
Tự chứng minh EC ⊥ BH
Lại có:
+) JM là đường TB của tam giác HBC (cmt) => JM // BH
+) IM là đường TB của tam giác EBC (cmt) => IM // EC
=>IM vuông góc với JM (**)
Từ (*) và (**) => đpcm
Đề bị sai từ câu đầu tiên. Tam giác chỉ có thể kí hiệu bằng 3 chữ cái in hoa.
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
a: TA có: \(\hat{BAG}=\hat{BAC}+\hat{GAC}=\hat{BAC}+90^0\)
\(\hat{EAC}=\hat{EAB}+\hat{BAC}=90^0+\hat{BAC}\)
Do đó: \(\hat{BAG}=\hat{EAC}\)
Xét ΔBAG và ΔEAC có
BA=EA
\(\hat{BAG}=\hat{EAC}\)
AG=AC
Do đó: ΔBAG=ΔEAC
=>BG=EC
Gọi O là giao điểm của BG và EC
ΔBAG=ΔEAC
=>\(\hat{ABG}=\hat{AEC}\)
Xét tứ giác AEBO có \(\hat{AEO}=\hat{ABO}\)
nên AEBO là tứ giác nội tiếp
=>\(\hat{BOE}=\hat{BAE}=90^0\)
=>BG⊥EC tại O
b: Q là tâm của hình vuông ABDE
=>Q là trung điểm chung của AD và BE
N là tâm của hình vuông ACFG
=>N là trung điểm chung của AF và CG
Xét ΔEBC có
Q,M lần lượt là trung điểm của BE,BC
=>QM là đường trung bình của ΔEBC
=>QM//EC và \(QM=\frac{EC}{2}\)
Xét ΔGEC có
P,N lần lượt là trung điêm của GE,GC
=>PN là đường trung bình cua ΔGEC
=>PN//EC và \(PN=\frac{EC}{2}\)
QM//EC
PN//EC
Do đó: QM//PN
\(QM=\frac{EC}{2}\)
\(PN=\frac{EC}{2}\)
Do đó: QM=PN
Xét ΔEBG có
Q,P lần lượt là trung điểm của EB,EG
=>QP là đường trung bình của ΔEBG
=>QP//BG và \(QP=\frac{BG}{2}\)
\(QP=\frac{BG}{2}\)
\(QM=\frac{EC}{2}\)
mà BG=EC
nên QP=QM
QP//BG
BG⊥EC
Do đó: QP⊥EC
QP⊥EC
EC//QM
Do đó: QP⊥QM
Xét tứ giác MNPQ có
MQ//NP
MQ=NP
Do đó: MNPQ là hình bình hành
Hình bình hành MNPQ có QM⊥QP
nên MNPQ là hình chữ nhật
Hình chữ nhật MNPQ có QM=QP
nên MNPQ là hình vuông