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a. xét 2 tam giác vuông AHB và ADH có
góc BAH _ chung
suy ra tam giác AHB đồng dạng với tam giác AHD (g.g)
suy ra AH/AD=AB/AH
suy ra AH2=AB.AD
~mình chỉ piết tới đó thôi nha
b. xét 2 tam giác vuông AED và ABC có
góc A chung
suy ra tam giác AED đồng dạng với tam giác ABC
suy ra AD/AC=AE/AB
suy ra AD.AB= AE.AC
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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!
Bạn tự vẽ hình
a/ Dễ thấy ADHE là hình chữ nhật vì góc A = góc E = góc D = 90 độ
=> góc ADE = góc AHE (t/c hình chữ nhật)
Mà góc AHE + góc EHC = 90 độ ; góc ECH + góc EHC = 90 độ
=> Góc AHE = góc ECH hay góc C = góc ADE
b/ Bạn tham khảo ở đây : http://olm.vn/hoi-dap/question/677639.html
a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có
góc B chung
Do đo:ΔABC đồng dạng với ΔHBA
c: Xét ΔAHB vuông tại H có HD là đường cao
nên \(AD\cdot AB=AH^2\left(1\right)\)
Xét ΔAHC vuông tại H có HE là đường cao
nên \(AE\cdot AC=AH^2\left(2\right)\)
Từ (1) và (2) suy ra \(AD\cdot AB=AE\cdot AC\)
hay AD/AC=AE/AB
\(S_{ABC}=\dfrac{AB\cdot AC}{2}=15\cdot40=600\left(cm^2\right)\)
DE=AH=24cm
Xét ΔADE vuông tại A và ΔACB vuông tại A có
AD/AC=AE/AB
Do đo: ΔADE đồng dạng với ΔACB
Suy ra: \(\dfrac{S_{ADE}}{S_{ACB}}=\left(\dfrac{DE}{CB}\right)^2=\left(\dfrac{24}{50}\right)^2=\dfrac{144}{625}\)
hay \(S_{ADE}=138.24\left(cm^2\right)\)
Cậu ơi, cậu hk lm câu c cho tớ hả :3?