Đáp án:

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

b) $AB.CA=CB.AH$

c) $BC=20cm, DB=\dfrac{60}{7}cm, DC=\dfrac{80}{7}cm$

d) $\dfrac{S_{\triangle ABD}}{S_{\triangle ACD}}=\dfrac{3}{4}$

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)

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

$\to\dfrac{AH}{AB}=\dfrac{CA}{CB}\\\to AB.CA=CB.AH$

c)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

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

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

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

Ta có:

$AB+AC=BC=20(cm)\\\to AB+\dfrac{4}{3}AB=20\\\to 7AB=60\\\to AB=\dfrac{60}{7}(cm)\to AC=\dfrac{4}{3}.\dfrac{60}{7}=\dfrac{80}{7}(cm)$

d)

$S_{\triangle ABD}=\dfrac{1}{2}.AH.DB\\S_{\triangle ACD}=\dfrac{1}{2}.AH.CD\\\to\dfrac{S_{\triangle ABD}}{S_{\triangle ACD}}=\dfrac{\dfrac{1}{2}.AH.DB}{\dfrac{1}{2}.AH.DC}=\dfrac{DB}{DC}=\dfrac{3}{4}$