Định lí đảo và hệ quả của định lí Ta - let

Áp dụng Pi-ta-go trong tam giác vuông $AMN$ ta có $MN=\sqrt{A{{M}^{2}}+A{{N}^{2}}}=\sqrt{{{8}^{2}}+{{6}^{2}}}=10$

Ta có $MN//BC\Rightarrow \dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{MN}{AM}$

$\begin{array}{l} \Leftrightarrow \dfrac{8}{AB}=\dfrac{6}{AC}=\dfrac{10}{12} \\ \Rightarrow AB=\dfrac{8.12}{10}=\dfrac{48}{5};AC=\dfrac{6.12}{10}=\dfrac{36}{5} \end{array}$

Khi đó diện tích tam giác ABC bằng: ${{S}_{ABC}}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.\dfrac{48}{5}.\dfrac{36}{5}=\dfrac{864}{25}$

Ta có: $AD=DE=EB=\dfrac{1}{3}AB\left( gt \right)\left( 1 \right)$ $\Rightarrow AE=AD+DE=\dfrac{2}{3}AB\left( 2 \right)$

Trong $\Delta ABC$ , ta có: $DM//BC\left( gt \right)$

Nên $\dfrac{AD}{AB}=\dfrac{DM}{BC}\Leftrightarrow \dfrac{AD}{AB}=\dfrac{DM}{a}\left( 3 \right)$

Từ $\left( 1 \right);\left( 3 \right)$ ta có $\dfrac{1}{3}=\dfrac{DM}{a}\Rightarrow DM=\dfrac{a}{3}$

Trong $\Delta ABC$ , ta có: $EN//BC\left( gt \right)$ $\Rightarrow \dfrac{AE}{AB}=\dfrac{EN}{BC}\Rightarrow \dfrac{AE}{AB}=\dfrac{EN}{a}\left( 4 \right)$

Từ $\left( 2 \right);\left( 4 \right)$ suy ra: $\dfrac{2}{3}=\dfrac{EN}{a}\Rightarrow EN=\dfrac{2a}{3}$

$\Rightarrow DM+EN=\dfrac{a}{3}+\dfrac{2a}{3}=a$

$\begin{array}{l} MN//BC\Rightarrow \dfrac{MA}{AC}=\dfrac{NA}{AB}=\dfrac{MN}{BC} \\ \Leftrightarrow \dfrac{12}{16}=\dfrac{y}{14}=\dfrac{13}{x} \\ \Rightarrow y=\dfrac{12.14}{16}=\dfrac{21}{2};x=\dfrac{13.16}{12}=\dfrac{52}{3} \end{array}$

$\Rightarrow x+y=\dfrac{21}{2}+\dfrac{52}{3}=\dfrac{167}{6}cm$

Vì $BK//AD$ và $AB//DG$ nên theo hệ quả định lý Ta-lét ta có:

$\dfrac{EK}{AE}=\dfrac{EB}{ED}=\dfrac{AE}{EG}\Rightarrow A{{E}^{2}}=EK.EG$

Ta có $MB//DN\Rightarrow \dfrac{OB}{OD}=\dfrac{MO}{ON}=\dfrac{1,5}{6}=\dfrac{1}{4}\Rightarrow NO=4MO$

Mà $MN = MO + ON = 5cm \Rightarrow OM + 4OM = 5cm \Rightarrow OM = 1cm$

$\Rightarrow ON=4cm$

Qua C kẻ $CF//AD,F\in AB$ , ta có: $\widehat{F}\,\,=\,\,\widehat{DAB}\,\,=\,\,{{60}^{0}}$ (1)

$\widehat{FCA}\,\,=\,\,\widehat{CAD}\,\,=\,\,{{60}^{0}}$ (2)

Từ (1) và (2) suy ra $\vartriangle AFC$ đều $\Rightarrow AF=FC=AC\Rightarrow BF=AB+AF=AB+AC$

Áp dụng hệ quả định lý Ta-lét vào $\vartriangle BFC,AD//FC:$

$\dfrac{\text{AD}}{\text{FC}}=\dfrac{\text{BA}}{\text{BF}}$ hay $\dfrac{\text{AD}}{\text{AC}}=\dfrac{\text{AB}}{\text{AB}+\text{AC}}$ $\Rightarrow \text{AD }=\dfrac{\text{AC}\text{.AB}}{\text{AB}+\text{AC}}$

$\Rightarrow \dfrac{\text{1}}{\text{AD}}=\dfrac{\text{AB}+\text{AC}}{\text{AB}\text{.AC}}=\dfrac{\text{1}}{\text{AB}}+\dfrac{\text{1}}{\text{AC}}$