Đáp án:

a) $\triangle AKB\backsim\triangle DKE$

b) $\dfrac{AK}{KD}=\dfrac{AC}{CE}$

c) $BD=\dfrac{45}{7}cm, DC=\dfrac{60}{7}cm, S_{ABDE}=\dfrac{1782}{49}cm^2$

Giải thích các bước giải:

a)

Xét $\triangle AKB$ và $\triangle DKE$:

$\widehat{AKB}=\widehat{DKE}$ (đối đỉnh)

$\widehat{ABK}=\widehat{DEK}$ (so le trong)

$\to\triangle AKB\backsim\triangle DKE$ (g.g)

b)

$\triangle AKB\backsim\triangle DKE$ (cmt)

$\to\dfrac{AK}{DK}=\dfrac{AB}{DE}$

$\triangle ABC$ có $DE//AB$ (gt)

$\to\dfrac{DE}{AB}=\dfrac{CE}{AC}$ (định lý Talet)

$\to\dfrac{AB}{DE}=\dfrac{AC}{CE}$

$\to\dfrac{AK}{DK}=\dfrac{AC}{CE}$

c)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{9^2+12^2}=15(cm)$

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

$\to\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{9}{12}=\dfrac{3}{4}\\\to DB=\dfrac{3}{4}DC\\to DB+DC=BC=15(cm)\\\to\dfrac{3}{4}DC+DC=15\\\to 7DC=60\\\to DC=\dfrac{60}{7}(cm)\to DB=\dfrac{3}{4}.\dfrac{60}{7}=\dfrac{45}{7}(cm)$

$\triangle ABC$ có $DE//AB$ (gt)

$\to\dfrac{AE}{AC}=\dfrac{BD}{BC}$ (định lý Talet)

$\to AE=\dfrac{AC.BD}{BC}=\dfrac{12.\dfrac{45}{7}}{15}=\dfrac{36}{7}(cm)$

$\to CE=AC-AE=12-\dfrac{36}{7}=\dfrac{48}{7}(cm)$

Ta có: $\dfrac{DE}{AB}=\dfrac{CE}{CA}$ (cmt)

$\to DE=\dfrac{AB.CE}{CA}=\dfrac{9.\dfrac{48}{7}}{12}=\dfrac{36}{7}(cm)$

$\to S_{ABDE}=\dfrac{(AB+DE).AE}{2}=\dfrac{\left(9+\dfrac{36}{7}\right).\dfrac{36}{7}}{2}=\dfrac{1782}{49}(cm^2)$

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