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Câu 1
a, Vì tứ giác ABCD là hình thang
⇒ AB // CD
ΔCOD có AB // CD
⇒ ΔAOB ~ ΔCOD
⇒ \(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\)(đpcm)
b, Vì AB // CD ⇒ AM // CN
ΔCON có AM // CN
⇒ ΔAOM ~ ΔCON
⇒ \(\frac{OA}{OC}=\frac{OM}{ON}\)
mà \(\frac{OA}{OC}=\frac{AB}{CD}\)(câu a)
⇒ \(\frac{OM}{ON}=\frac{AB}{CD}\)
⇒ \(\frac{OM}{AB}=\frac{ON}{CD}\) (đpcm)
Câu 2
a, Vì ΔABC vuông tại A
⇒ \(\widehat{BAC}=90^0\)
Vì AH là đường cao của ΔABC
⇒ AH ⊥ BC
⇒ \(\widehat{H_1}=\widehat{H_2}=90^0\)
ΔABC và ΔHBA có
\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{H_1}=90^0\\\widehat{ABC}chung\end{matrix}\right.\)
⇒ ΔABC ~ ΔHBA (g.g)
⇒ \(\frac{AB}{HB}=\frac{BC}{AB}\) (1)
⇒ AB2 = BH . BC (đpcm)
b, ΔABC có BF là đường phân giác
⇒ \(\frac{BC}{AB}=\frac{FC}{FA}\) (2)
ΔABH có HE là đường phân giác
⇒ \(\frac{AB}{HB}=\frac{AE}{EH}\)(3)
Từ (1), (2), (3) ⇒ \(\frac{AE}{EH}=\frac{FC}{FA}\)
⇒ \(\frac{EH}{EA}=\frac{FA}{FC}\) (đpcm)
Chúc b
ạn học tốt !!
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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: XétΔABC vuông tại A và ΔHBA vuông tại H có
góc B chung
Do đó: ΔABC\(\sim\)ΔHBA
Suy ra: BA/BH=BC/BA
hay \(BA^2=BH\cdot BC\)
b: Xét ΔBAD có MN//AD
nên MN/AD=BM/BA(1)
Xét ΔBCA có MH//AC
nên MH/AC=BM/BA(2)
Từ (1) và (2) suy ra MN/AD=MH/AC
hay MN/MH=AD/AC
Bài 1
A B C M H K 1 a, Xét ΔABM và ΔACB có
\(\left\{{}\begin{matrix}\widehat{BAC}\text{ chung}\\\widehat{ABM}=\widehat{C}\text{(gt)}\end{matrix}\right.\)
⇒ ΔABM ~ ΔACB (g.g)(đpcm)
b, Vì ΔABM ~ ΔACB
⇒ \(\frac{AB}{AC}=\frac{AM}{AB}\)
⇒ AB2 = AM . AC
⇒ AM = \(\frac{AB^2}{AC}=\frac{2^2}{4}=\frac{4}{4}=1\) (cm)
Vậy AM = 1cm
c, Vì ΔABM ~ ΔACB
⇒ \(\widehat{M_1}=\widehat{ABC}\)
⇒ \(\widehat{M_1}=\widehat{ABH}\)
Vì AH ⊥ BC ⇒ \(\widehat{AHB}=90^0\)
AK ⊥ BM ⇒ \(\widehat{AKM}=90^0\)
ΔAHB và ΔAKM có
\(\left\{{}\begin{matrix}\widehat{ABH}=\widehat{M_1}\\\widehat{AHB}=\widehat{AKM}=90^0\end{matrix}\right.\)
⇒ ΔAHB ~ ΔAKM (g.g)
⇒ \(\frac{AB}{AM}=\frac{AH}{AK}\)
⇒ AB . AK = AH . AM (đpcm)
d, Vì ΔABH ~ ΔAMK
⇒ \(\frac{\text{SΔABH}}{\text{SΔAMK}}=\left(\frac{AB}{AM}\right)^2\) (Tỉ số diện tích của 2 tam giác đồng dạng bằng bình phương tỉ số đồng dạng)
⇒ \(\frac{\text{SΔABH}}{\text{SΔAMK}}=\left(\frac{2}{1}\right)^2\)
⇒ \(\frac{\text{SΔABH}}{\text{SΔAMK}}=4\)
⇒ SΔABH = 4SΔAMK (đpcm)
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=>\(\frac{FC}{AF}=\frac{AE}{EH}=>\frac{EH}{AE}=\frac{AF}{FC}\)