Bạn tử kẻ hình nhé .

a)\(\Delta ABD~\Delta ACE\left(g.g\right)\)

\(\Rightarrow\frac{AB}{AC}=\frac{AD}{AE}\)

\(\Rightarrow\Delta ADE~\Delta ABC\left(c.g.c\right)\)

\(\Rightarrow\frac{S_{ADE}}{S_{ABC}}=\left(\frac{AD}{AB}\right)^2=cos^2\widehat{BAC}\)

\(\Rightarrow S_{ADE}=S_{ABC}.cos^2\widehat{BAC}\)

b)Ta có : \(S_{BCDE}=S_{ABC}-S_{ADE}=S_{ABC}-S_{ABC}.cos^2\widehat{BAC}=S_{ABC}\left(1-cos^2\widehat{BAC}\right)=S_{ABC}.sin^2\widehat{BAC}\)

a) Ta có: \(cosA=\dfrac{AD}{AB};cosA=\dfrac{AE}{AC}\)

Do đó: \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

Vậy \(\Delta ADE\sim\Delta ABC\left(c-g-c\right)\) do đó

\(\dfrac{S_{ADE}}{S_{ABC}}=\left(\dfrac{AD}{AB}\right)^2=cos^2A\)

Suy ra: \(S_{ADE}=S_{ABC}.cos^2A\)

b) \(S_{BCDE}=S_{ABC}-S_{ADE}=S_{ABC}-S_{ABC}.cos^2A\)

\(=S_{ABC}\left(1-cos^2A\right)=S_{ABC}sin^2A\)

a, \(\bigtriangleup{ABD} \sim \bigtriangleup{ACE}\) (g.g)

\(\Rightarrow\) \(\dfrac{AB}{AC} = \dfrac{AD}{AE}\) \(\Rightarrow\) \(\dfrac{AB}{AD} = \dfrac{AC}{AE}\)

\(\Rightarrow\) \(S_{ABC} \sim S_{ADE}\) (c.g.c)

\(\Rightarrow\) \(\dfrac{S_{ADE}}{S_{ABC}} = k^2 = ({\dfrac{AD}{AB}})^2\) = \(cos^2A\)

\(\Rightarrow\) \(S_{ADE} = S_{ABC} . cos^2A\) (đpcm)

b, \(S_{BCDE} = S_{ABC} - S_{ADE}\)

\(= S_{ABC} - S_{ABC} . cos^2A \)

= \(S_{ABC} (1-cos^2A)\)

= \(S_{BCDE} = S_{ABC} . sin^2A \) (đpcm)

Xét tứ giác BEDC có góc BEC=góc BDC=90 độ

nên BEDClà tứ giác nội tiếp

=>góc AED=góc ACB

=>ΔAED đồng dạng với ΔACB

Suy ra: \(\dfrac{S_{ADE}}{S_{ABC}}=\left(\dfrac{AD}{AB}\right)^2=cos^2A\)

hay \(S_{ADE}=S_{ABC}\cdot cos^2A\)

Xét ΔADB vuông tại D và ΔAEC vuông tại E có

góc BAD chung

DO đó ΔADB đồng dạng với ΔAEC

Suy ra: AD/AE=AB/AC

=>AD/AB=AE/AC

=>ΔADE đồng dạng với ΔABC

=>\(\dfrac{S_{ADE}}{S_{ABC}}=\left(\dfrac{AD}{AB}\right)^2=cos^2A\)

hay \(S_{ADE}=S_{ABC}\cdot cos^2A\)

Xét tứ giác BEDC có góc BEC=góc BDC=90 độ

nên BEDC là tứ giác nội tiếp

=>góc AED=góc ACB

=>ΔAED đồng dạng với ΔACB

=>\(\dfrac{S_{ADE}}{S_{ACB}}=\left(\dfrac{AD}{AC}\right)^2=cos^2A\)

hay \(S_{ADE}=S_{ABC}\cdot cos^2A\)