Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Xét tam giác ABC và tam giác HBA có Góc ABC chungg,góc BHA=góc BAC=90 độ
=> Tam giác ABC đồng dạng với tam giác HBA(gg)=> \(\frac{AB}{HB}=\frac{BC}{AB}\)=> AB^2=BH.BC
b)Tam giác ABC có BF là phân giác góc ABC=>\(\frac{BC}{AB}=\frac{FC}{AF}\)mà \(\frac{AB}{HB}=\frac{BC}{AB}\)=>\(\frac{AB}{BH}=\frac{FC}{AF}\left(1\right)\)
Tam giác ABH có BE là phân giác goc ABH =>\(\frac{BA}{BH}=\frac{AE}{EH}\left(2\right)\)
Từ 1 và 2=>\(\frac{FC}{AF}=\frac{AE}{EH}=>\frac{EH}{AE}=\frac{AF}{FC}\)
Theo câu a) ta có: \(AH^2=AI.AB\left(1\right)\)
Xét tam giác AHK và tam giác ACH có:
góc A chung; góc AKH = góc AHC = 900
=> tam giác AHK đồng dạng với tam giác ACH (g-g)
=>\(\dfrac{AK}{AH}=\dfrac{AH}{AC}\Rightarrow AK.AC=AH^2\left(2\right)\)
Từ (1)(2) => \(AI.AB=AK.AC\Rightarrow\dfrac{AI}{AC}=\dfrac{AK}{AB}\)
Xét tam giác AIK và tam giác ABC có:
góc A chung; \(\dfrac{AI}{AC}=\dfrac{AK}{AB}\)
=> Tam giác AIK đồng dạng với tam giác ACB (c-g-c)
a) Xét tam giác AIH và tam giác AHB có:
góc BAH chung; góc AIH = góc AHB (= 900)
=> tam giác AIH = tam giác AHB (g-g)
\(\Rightarrow\dfrac{AH}{AI}=\dfrac{AB}{AH}\Rightarrow AH^2=AI.AB\)
Bài 3:
a: Xét ΔHBA vuông tại H và ΔABC vuông tại A có
góc HBA chung
DO đó: ΔHBA\(\sim\)ΔABC
SUy ra: BA/BC=BH/BA
hay \(BA^2=BH\cdot BC\)
b: \(BC=\sqrt{12^2+16^2}=20\left(cm\right)\)
Xét ΔABC có AD là phân giác
nên BD/AB=CD/AC
=>BD/3=CD/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{20}{7}\)
Do đó: BD=60/7(cm); CD=80/7(cm)
ttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt
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!