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ABCHKIEF
a)
Xét \(\Delta\)ABC và \(\Delta\)HBA có:
^BAC = ^BHA ( = 90 độ )
^ABC = ^HBA ( ^B chung )
=> \(\Delta\)ABC ~ \(\Delta\)HBA
b) AB = 3cm ; AC = 4cm
Theo định lí pitago ta tính được BC = 5 cm
Từ (a) => \(\frac{AB}{BH}=\frac{BC}{AB}\Rightarrow BH=\frac{AB^2}{BC}=1,8\)m
c) Xét \(\Delta\)AHC và \(\Delta\)AKH có: ^AKH = ^AHC = 90 độ
và ^HAC = ^HAK ( ^A chung )
=> \(\Delta\)AHC ~ \(\Delta\)AKH
=> \(\frac{AH}{AK}=\frac{AC}{AH}\Rightarrow AH^2=AC.AK\)
d) Bạn kiểm tra lại đề nhé!
a: Ta có: ΔBKC vuông tại K
mà KM là trung tuyến
nên KM=BC/2
Ta có: ΔBHC vuông tại H
mà HM là trung tuyến
nên HM=BC/2
=>HM=KM
b: KẻMN vuông góc với HK
Vì ΔMHK cân tại M có MN là đường cao
nên N là trung điểm của HK
Xét hình thang BDEC có
M là trung điểm của B
MN//BD//EC
DO đó:N là trung điểm của DE
=>DN=NE
=>DK=HE
+ Xét hai tg vuông BKC và tg vuông CHB có
Cạnh huyền BC chung (1)
\(S_{ABC}=\frac{AB.CK}{2}=\frac{AC.BH}{2}\) Mà AB=AC => BH=CK (2)
Từ (2) Và (2) => tg BKC = tg CHB (cạnh huyền và cạnh góc vuông tương ứng bằng nhau) => BK=CH (*)
Mà AB=AC=AK+BK=AH+CH => AK=AH => tg AKH cân tại A
+ Xét tg cân AKH có
^AKH=^AHK=(180-^BAC)/2 (3)
+ Xét tg cân ABC có
^ABC=^ACB=(180-^BAC)/2 (4)
Từ (3) và (4) => ^AKH=^ABC => KH//BC (có hai góc đồng vị bằng nhau) (**)
Từ (*) và (**) => BKHC là hình thang cân
a) Xét \(\Delta\) ABH và \(\Delta\)ACK
Ta có: Góc A chung
AB = AC
góc AHB = góc AKC ( =90o )
=> \(\Delta\)AKC = \(\Delta\)AHB ( ch-gn)
=> BH = CK
=> AK = AH
=>\(\dfrac{AK}{KB}\) = \(\dfrac{AH}{HB}\)
=> HK // BC
b) Xét \(\Delta\)IAC và \(\Delta\)HBC
Ta có : Góc I = Góc H (=900)
Góc C chung
=> \(\Delta\)IAC \(\infty\) \(\Delta\)HBC (g.g)
Xét \(\Delta\)AKH và \(\Delta\)ABC
Ta có:
\(\dfrac{AK}{AB}\) = \(\dfrac{AH}{AC}\) ( do HK // BC )
Góc A chung
=> \(\Delta\)AKH \(\infty\) \(\Delta\)ABC (c.g.c)
c) Tự thay vào làm nhé!!
đoạn AB lon hon nha ban
đầu bài đúng!
SABC=BH.AC/2 SABC=CK.AB/2 Suy ra BH.AC=CK.AB => AC/AB=CK/BH.
Do AC>AB nên AC/AB>1 dẫn tới CK/BH>1
Kết luận: CK>BH (đpcm)