a/ Xét $ΔDEC$ và $ΔABC$:
$\widehat C:chung$
$\widehat{EDC}=\widehat{BAC}(=90^\circ)$
$→ΔDEC\backsim ΔABC(g-g)$
$→\dfrac{DE}{DC}=\dfrac{AB}{AC}$
mà $\dfrac{AB}{AC}=\dfrac{3}{4}$
$→\dfrac{DE}{DC}=\dfrac{3}{4}$
b/ $AD$ là đường phân giác $\widehat A$ trong $ΔABC$
$→\dfrac{AB}{AC}=\dfrac{DB}{DC}$
mà $\dfrac{AB}{AC}=\dfrac{3}{4}$
$→\dfrac{DB}{DC}=\dfrac{3}{4}$
mà $\dfrac{DE}{DC}=\dfrac{3}{4}$
$→\dfrac{DB}{DC}=\dfrac{DE}{DC}\left(=\dfrac{3}{4}\right)$
$→DB=DE$
c/ $\dfrac{AB}{AC}=\dfrac{3}{4}$
$↔AB=\dfrac{3AC}{4}$
Áp dụng định lý Pytago vào $ΔABC$ vuông tại $A$:
$AB^2+AC^2=BC^2\\↔\left(\dfrac{3AC}{4}\right)^2+AC^2=20^2\\↔\dfrac{9AC^2}{16}+AC^2=400\\↔\dfrac{9AC^2}{16}+\dfrac{16AC^2}{16}=\dfrac{6400}{16}\\↔9AC^2+16AC^2=6400\\↔25AC^2=6400\\↔AC^2=256\\↔AC=16cm(AC>0)\\→AB=\dfrac{3.16}{4}=12(cm)$
Vậy $AB=12cm,AC=16cm$
d/ Xét $ΔABH$ và $ΔCBA$:
$\widehat B:chung$
$\widehat{AHB}=\widehat{CAB}(=90^\circ)$
$→ΔABH\backsim ΔCBA(g-g)$
$→\dfrac{AB}{AH}=\dfrac{BC}{AC}$
$↔AB.AC=AH.BC$ hay $12.16=AH.20$
$↔192=AH.20\\↔9,6cm=AH$
Vậy $AH=9,6cm$
