Cho tam giác ABC vuông tại A. Dựng về phía ngoài tam giác ấy hình vuông BCEF. Gọi I là tâm hình vuông. Tính AI biết AB = 6 , AC = 8
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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)
=> \(\Delta O_1DO_2=\Delta ADO_3\left(c-g-c\right)\) => \(O_1O_2=AO_3\)
Ta có: \(\widehat{DCB}=\widehat{CBA}\left(=45^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên AB//CD
Xét tứ giác ABDC có DC//BA
nên ABDC là hình thang
mà \(\widehat{CAB}=90^0\)
nên ABDC là hình thang vuông
1. ∠DAE =∠BAD + ∠DAE + ∠CAE = 45⁰ + 90⁰ + 45⁰ = 180⁰
=> D, A, E thẳng hàng
2. △ABC vuông ở A ; I là trung điểm của BC
=> IA = BC/2
=> IA = IC
=> I ∈ trung trực của AB
△ABD vuông cân ở D => DA = DB
=> D ∈ trung trực của AB
=> DI là ∈ trung trực của AB
=> DI ⊥ AB
tương tự IE ⊥ AC
3. F là gì nhỉ ?
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.