Cho \(\Delta ABC\) vuông tại A, đường cao AH.
a) CM: \(\Delta ABC\)đồng dạng \(\Delta HBA\)
b) Tính BH? biết AB=3cm, AC=4m
c) Kẻ HK vuông góc với AC. CM: \(\Delta AHC\)và \(\Delta AKH\)đồng dạng, từ đó => \(AH^2=AK.AC\)
d) Kẻ HI vuông góc với AB, Các tia HI, HK cắt 1 đường thẳng a bất kì qua A lần lượt tại E, F. Chứng minh \(\frac{CK}{FI}=\frac{KE}{IA}\)
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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!
