a/ Áp dụng định lý Pytago vào $ΔABC$ vuông tại $A$
$→BC=\sqrt{AB^2+AC^2}=\sqrt{81+144}=\sqrt{225}=15$ (cm)
$AD$ là đường phân giác $\widehat{A}$
$→\dfrac{AB}{AC}=\dfrac{DB}{DC}$
$→\dfrac{9}{12}=\dfrac{DB}{DC}$ hay $\dfrac{3}{4}=\dfrac{DB}{DC}$
$→\dfrac{DB}{3}=\dfrac{DC}{4}=\dfrac{DB+DC}{3+4}=\dfrac{15}{7}$
$→\begin{cases}DB=\dfrac{45}{7}\\DC=\dfrac{60}{7}\end{cases}$
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b/ Xét $ΔABC$ và $ΔEDC$:
$\widehat{C}$: chung
$\widehat{BAC}=\widehat{DEC}$ ($=90^\circ$)
$→ΔABC\backsim ΔEDC(g-g)$
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c/ $ΔABC\backsim ΔEDC$
$→\dfrac{AB}{AC}=\dfrac{ED}{EC}$ mà $\dfrac{AB}{AC}=\dfrac{DB}{DC}$
$→\dfrac{ED}{EC}=\dfrac{DB}{DC}$
$→EC.BD=ED.DC$
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d/ Kẻ đường cao AH ứng BC
$S_{ΔABD}=\dfrac{1}{2}.AH.BD$
$S_{ΔACD}=\dfrac{1}{2}.AH.CD$
$→\dfrac{S_{ΔABD}}{S_{ΔACD}}=\dfrac{\dfrac{1}{2}.AH.BD}{\dfrac{1}{2}.AH.CD}=\dfrac{BD}{CD}=\dfrac{3}{4}$
