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bài đó mình cũng biết làm nhưng dài lắm nếu bn muốn biêt mình gợi ý cho
Bài này dài dòng lắm bạn ạ viết cũng phải chết mỏi
Ủng hộ nha
a: Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\dfrac{BH}{CH}=\dfrac{AB^2}{AC^2}\)
b: \(\dfrac{BE}{CF}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}\)
\(=\dfrac{BH^2}{CH^2}\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)
e: \(BE\cdot CF\cdot BC\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\cdot BC\)
\(=\dfrac{AH^4}{AB\cdot AC}\cdot BC=\dfrac{AH^4}{AH\cdot BC}\cdot BC=AH^3\)
\(=EF^3\)
a: \(\dfrac{AB^2}{AC^2}=\dfrac{BH\cdot CB}{CH\cdot BC}=\dfrac{BH}{CH}\)
b: \(\dfrac{BD}{CE}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}\)
\(=\dfrac{BH^2}{CH^2}\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)
c: \(BD\cdot CE\cdot BC\)
\(=\dfrac{BH^2}{AB}\cdot\dfrac{CH^2}{AC}\cdot BC\)
\(=\dfrac{AH^4}{AH}=AH^3=DE^3\)
a: \(BD\cdot CE\cdot BC\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\cdot\dfrac{AB\cdot AC}{AH}\)
\(=\dfrac{AH^4}{AH}=AH^3\)
b: \(\dfrac{BD}{CE}=\dfrac{HB^2}{AB}:\dfrac{HC^2}{AC}=\dfrac{HB^2}{AB}\cdot\dfrac{AC}{HC^2}=\dfrac{AB^4}{AB}\cdot\dfrac{AC}{AC^4}=\dfrac{AB^3}{AC^3}\)
a) + ΔADB ∼ ΔAEC ( g.g )
\(\Rightarrow\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow\frac{AD}{AE}=\frac{AB}{AC}\)
+ ΔADE ∼ ΔABC ( c.g.c )
b) + AC // MH \(\Rightarrow\frac{AH}{AB}=\frac{MC}{CB}\)
+ AB // MK \(\Rightarrow\frac{CK}{AC}=\frac{MC}{CB}\)
\(\Rightarrow\frac{CK}{AC}-\frac{AH}{AB}=0\)
\(\Rightarrow\left(\frac{CK}{AC}+1\right)-\frac{AH}{AB}=1\)
\(\Rightarrow\frac{AK}{AC}-\frac{AH}{AB}=1\)
a) Ta có: \(\left(\dfrac{AB}{AC}\right)^2=\dfrac{AB^2}{AC^2}=\dfrac{BH.BC}{CH.BC}=\dfrac{BH}{HC}\)
b) Ta có: \(\left(\dfrac{CA}{AB}\right)^4=\left(\dfrac{CA^2}{AB^2}\right)^2=\left(\dfrac{CH.BC}{BH.BC}\right)^2=\dfrac{CH^2}{BH^2}=\dfrac{CE.CA}{BD.BA}\)
\(=\dfrac{CE}{BD}.\dfrac{CA}{BA}\Rightarrow\left(\dfrac{CA}{AB}\right)^3=\dfrac{CE}{BD}\)
c) Ta có: \(AH^4=\left(AH^2\right)^2=\left(BH.CH\right)^2=BH^2.CH^2\)
\(=BD.BA.CE.CA=BD.CE\left(AB.AC\right)=BD.CE.AH.BC\)
\(\Rightarrow BD.CE.BC=AH^3\)
d) Vì \(\angle HDA=\angle HEA=\angle DAE=90\Rightarrow ADHE\) là hình chữ nhật
\(\Rightarrow AH=DE\Rightarrow AH^2=DE^2=DH^2+HE^2\)
Ta có: \(3AH^2+BD^2+CE^2=2AH^2+\left(DH^2+BD\right)^2+\left(HE^2+CE^2\right)\)
\(=2.HB.HC+BH^2+CH^2=\left(BH+CH\right)^2=BC^2\)
Bạn ơi chỉ thêm cho mik câu b vs ạ