cho tam gaics abc nhọn aa',bb',cc'là đường cao của tam giác abc cắt nhau tại h . chứng minh rằng ha'/ha + hb'/hb +hc'/hc >= 3/2
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Những câu hỏi liên quan
11 tháng 8 2018
Bài 1:
Vì BH, CK vuông góc với HK nên BCHK là hình thang
Lấy P là trung điểm của BC
DP = BP (DP là đường trung tuyến của tam giác vuông DBC)
EP = BP( EP là đường trung trực của tam giác vuông BEC)
=> EP = DP
Vậy tam giác EMD là tam giác cân
Lấy Q là trung điểm của ED
Nên QP vuông góc với HK
Mà BP = CP
=> HQ = KQ
Hay EH + EQ = QD + DK mà EQ = QD
=> EH = DK
15 tháng 6 2022
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28 tháng 2 2017
a) Xét \(\Delta\) BHC vuông tại H và \(\Delta\) MHC vuông tại H có:
BH = HM (gt)
chung HC
=> \(\Delta\) BHC = \(\Delta\) MHC (cgv - cgv )
=> đpcm
b) Theo câu a : \(\Delta\) BHC = \(\Delta\) MHC
=> BC = CM (cặp cạnh tương ứng )
và \(\widehat{BCH}=\widehat{MCH}\) (cặp góc tương ứng )
Xét \(\Delta\) ABC và \(\Delta\) AMC có :
BC = MC (chứng minh trên )
\(\widehat{BCA}=\widehat{MCA}\) (chứng minh trên )
chung AC
=> \(\Delta\) ABC = \(\Delta\) AMC (c-g-c )
=> đpcm
28 tháng 2 2017
a) Xét \(\Delta BHC\) vuông tại H và \(\Delta MHC\) vuông tại H có:
CH chung
BH = MH (gt)
\(\Rightarrow\Delta BHC=\Delta MHC\left(cgv-cgv\right)\)
b) Xét \(\Delta ABH\) và \(\Delta AMH\) cùng vuông tại H có:
AH chung
BH = MH (gt)
\(\Rightarrow\Delta ABH=\Delta AMH\left(cgv-cgv\right)\)
\(\Rightarrow AB=AM\)
Do \(\Delta BHC=\Delta MHC\)
\(\Rightarrow BC=MC\)
Xét \(\Delta ABC\) và \(\Delta AMC\) có:
AB = AM (c/m trên)
AC chung
BC = MC (c/m trên)
\(\Rightarrow\Delta ABC=\Delta AMC\left(c.c.c\right)\)
T
3
4 tháng 2 2017
Định lí đảo Py-ta-go:
Trong một tam giác có tổng bình phương của hai cạnh bằng bình phương cạnh còn lại thì tam giác đó là tam giác vuông.
Xét tam giác ABC, ta có: AB2 + BC2 = 62 + 82 = 100
và AC2 = 102 = 100
=> tam giác ABC là tam giác vuông tại B.
Q
2
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
Lời giải:
Từ $A$ kẻ $AH\perp BC$.
Xét tam giác $ABH$: $\frac{AH}{BH}=\tan B$
$\Rightarrow BH=\frac{AH}{\tan B}=\frac{AH}{\tan 50^0}$
Xét tam giác $ACH$: $\frac{AH}{CH}=\tan C$
$\Rightarrow CH=\frac{AH}{\tan C}=\frac{AH}{\tan 30^0}$
Do đó:
$BC=BH+CH=AH(\frac{1}{\tan 50^0}+\frac{1}{\tan 30^0})$
$10=AH(\frac{1}{\tan 50^0}+\frac{1}{\tan 30^0})$
$AH=3,89$ (cm)
$S_{ABC}=\frac{AH.BC}{2}=\frac{3,89.10}{2}=19,45$ (cm vuông)
aa',bb',cc' chắc là đường cao à bạn
mình nhầm xíu mình chỉnh rồi đó bạn giúp mình với