Đáp án:

a) $\triangle HBA\backsim\triangle ABC$

b) $BC=20cm; AH=9,6cm; BH=7,2cm$

c) $DB≈8,6cm; DC≈11,4cm$

Giải thích các bước giải:

a)

Xét $\triangle HBA$ và $\triangle ABC$:

$\widehat{AHB}=\widehat{CAB}\,\,\,(=90^o)$

$\widehat{B}$: chung

$\to\triangle HBA\backsim\triangle ABC$ (g.g)

b)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+16^2}=20(cm)$

$\triangle HBA\backsim\triangle ABC$ (cmt)

$\to\dfrac{AH}{AB}=\dfrac{CA}{CB}\\\to AH=\dfrac{AB.CA}{CB}=\dfrac{12.16}{20}=9,6(cm)$

$\to\dfrac{BH}{AB}=\dfrac{BA}{CB}\\\to BH=\dfrac{AB^2}{CB}=\dfrac{12^2}{20}=7,2(cm)$

c)

$\triangle ABC$ có đường phân giác AD (gt)

$\to\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{12}{16}=\dfrac{3}{4}\\\to DB=\dfrac{3}{4}DC$

Ta có:

$DB+DC=BC=20(cm)\\\to\dfrac{3}{4}DC+DC=20\\\to 7DC=80\\\to DC≈11,4(cm)\to DB≈8,6(cm)$