Cho tam giác ABC nhọn có góc B>góc C ;2 đường cao AD và BE cắt nhau tại H.
a)C/m HC>HB
b)Vẽ HF vuông góc với AB;C/m C,H,F thẳng hàng.
c)C/m AB+AC>2AD
d)C/m HA+HB+HC <AB+AC
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Xét tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)( tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-70^0=110^0\)
Xét tam giác ABC có:
\(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=110^0\\\widehat{B}-\widehat{C}=40^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+40^0\right):2=75^0\\\widehat{C}=\left(110^0-40^0\right):2=35^0\end{matrix}\right.\)
Ta có: \(\widehat{C}< \widehat{A}< \widehat{B}< 90^0\)
Vậy tam giác ABC là tam giác nhọn
\(\Delta ABC\)có \(\widehat{A}=60^o\)\(\Rightarrow\widehat{B}+\widehat{C}=120^o\)
mà \(\widehat{B}=3.\widehat{C}\)\(\Rightarrow4.\widehat{C}=120^o\)\(\Rightarrow\widehat{C}=30^o\)\(\Rightarrow\widehat{B}=90^o\)
\(\Rightarrow\Delta ABC\)vuông tại B
vì góc C tù nên góc C lớn hơn 90 độ
Nên góc A, B chỉ có thể là góc nhọn, không thể là góc vuông vì một tam giác có tổng là 180 độ
kbn nha
Tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180\)
Mà \(\widehat{C}\)> 90
\(\Rightarrow\widehat{A}+\widehat{B}< 90\)
\(\Rightarrow\widehat{A}< 90;\widehat{B}< 90\)
Do B tù nên ta có góc ngoài của đỉnh B là góc nhọn, suy ra các góc A, C nhọn.
Câu 1:
Xét tam giác ABH vuông tại H, ta có:
AB2 = AH2 + HB2 (định lý Py-ta-go)
202 = AH2 + 162
400 = AH2 + 256
AH2 = 400 - 256
AH2 = 144
AH = \(\sqrt{144}\)= 12 (cm)
Xét tam giác AHC vuông tại H, ta có:
AC2 = AH2 + HC2 (định lý Py-ta-go)
AC2 = 122 + 52
AC2 = 144 + 25
AC2 = 169
AC = \(\sqrt{169}\)= 13 (cm)
Vậy AH = 12 cm
AC = 13 cm
Bài 2:
Xét tam giác AHC vuông tại H, ta có:
AC2 = AH2 + HC2 (định lý Py-ta-go)
152 = AH2 + 92
225 = AH2 + 81
AH2 = 225 - 81
AH2 = 144
AH = \(\sqrt{144}\)= 12 (cm)
Xét tam giác AHB vuông tại, ta có:
AB2 = AH2 + HB2 (định lý Py-ta-go)
AB2 = 122 + 52
AB2 = 144 + 25
AB2 = 169
AB = \(\sqrt{169}\)= 13 (cm)
Vậy AB = 13 cm
a: \(\widehat{B}>\widehat{C}\)
nên AB<AC
Xét ΔABC có AB<AC
mà DB là hình chiếu của AB trên BC
và DC là hình chiếu của AC trên BC
nên DB<DC
Xét ΔHBC có DB<DC
mà DB là hình chiếu của HB trên BC
và DC là hình chiếu của HC trên BC
nên HB<HC
b: Xét ΔABC có
AD là đường cao
BE là đường cao
AD cắt BE tại H
DO đó: H là trực tâm
=>CH\(\perp\)AB
mà HF\(\perp\)AB
và CH,HF có điểm chung là H
nên C,H,F thẳng hàng