[toán 8] hình khó

R

ronaldover7

Ta có BD là dường cao kẻ từ B
\Rightarrow $BA^2$-$AD^2$=$BC^2$-$DC^2$
\Rightarrow $BA^2$-$BC^2$=$AD^2$-$DC^2$
\Rightarrow $a^2$-$b^2$=(AD+DC)(AD-DC)
\Rightarrow $a^2$-$b^2$=b(AD-DC)
\Rightarrow AD-DC=$\frac{a^2-b^2}{b}$
\Rightarrow AD-DC+(AD+DC)=$\frac{a^2-b^2}{b}$+b
\Rightarrow 2AD=$\frac{a^2}{b}$
\Rightarrow AD=$\frac{a^2}{2b}$
ÁP dụng dl talet có ED//BC(tự cm nha):
$\frac{ED}{BC}$=$\frac{AD}{AC}$
\Rightarrow $\frac{ED}{b}$=$\frac{\frac{a^2}{2b}}{a}$

\Rightarrow ED= $\frac{\frac{a^2.b}{2b}}{a}$ = $\frac{a^2}{2a}$=$\frac{a}{2}$
 
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