Định nghĩa hai tam giác bằng nhau

$\begin{array}{l} \widehat{A}:\widehat{B}:\widehat{C}=2:3:4\Rightarrow \dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+4}=\dfrac{{{180}^{0}}}{9}={{20}^{0}}. \\ \Rightarrow \widehat{A}={{40}^{0}};\,\,\widehat{B}={{60}^{0}};\,\,\widehat{C}={{80}^{0}}. \end{array}$

$\Delta ACE$ có: $\widehat{A}+\widehat{ACE}={{90}^{0}}$ (Vì $CE\bot AB$ ).

$\widehat{HCD}$ có: ${{\widehat{H}}_{1}}+\widehat{ACE}={{90}^{0}}$ (Vì $BD\bot AC$ ).

$\Rightarrow \widehat{A}={{\widehat{H}}_{1}}.$

Ta lại có: ${{\widehat{H}}_{1}}+{{\widehat{H}}_{2}}={{180}^{0}}$ (kề bù).

Do đó: $\widehat{A}+{{\widehat{H}}_{2}}={{180}^{0}}.$

Ta có: $\widehat{xBA}+\widehat{yCA}=\left( \widehat{xBC}-\widehat{ABC} \right)+\left( \widehat{yCB}-\widehat{ACB} \right)$ $=\left( \widehat{xBC}+\widehat{yCB} \right)-\left( \widehat{ABC}+\widehat{ACB} \right)$ . (1)

Vì Bx và Cy cùng vuông góc với BC nên $\widehat{xBC}+\widehat{yCB}={{90}^{0}}+{{90}^{0}}={{180}^{0}}.$ (2)

Vì $\Delta ABC$ vuông tại A $\Rightarrow \widehat{ABC}+\widehat{ACB}={{90}^{0}}$ . (3)

Từ (1) (2) và (3) $\Rightarrow \,\,\,\widehat{xBA}+\widehat{yCA}={{180}^{o}}-{{90}^{0}}={{90}^{o}}$ .

Theo đề bài, ta có: $\widehat{B} < \widehat{A}.$ (1)

Hiển nhiên: $\widehat{B} < \widehat{B}+\widehat{C}.$ (2)

Cộng theo từng về (1) và (2), ta có: $2\widehat{B} < \widehat{A}+\widehat{B}+\widehat{C}={{180}^{0}}.$

Vậy $\widehat{B} < {{90}^{0}}.$

$\Delta MNP=\Delta ABC$

$\Rightarrow \,\,MN=AB\,;\,MP=AC\,;\,NP=BC$

Mà $MN=6cm\,;AC=7cm\,;NP=10cm$

$\Rightarrow \,\,AB=6cm,\,\,MP=7cm,\,BC=10cm$ .

$\Delta ABC=\Delta DEF\,$

$\Rightarrow \,\,AB=DE\,=6cm\,\,;$ $AC=DF=14cm\,;$ $BC=EF=10cm.$

Chu vi tam giác DEF là: $DE+DF+EF=6+14+10=30\,(cm).$

Ta có $\Delta ABC=\Delta D\text{EF}\Rightarrow \left\{ \begin{array}{l} \widehat{A}=\widehat{D} \\ \widehat{B}=\widehat{E} \\ \widehat{C}=\widehat{F} \end{array} \right.$

Mặt khác $\widehat{A}+\widehat{B}+\widehat{C}=\widehat{A}+\widehat{E}+\widehat{C}={{180}^{o}}\Rightarrow \widehat{C}={{60}^{o}}$

Vì $\Delta EFM=\Delta KPN$ nên $EF=KP\,;\,EM=KN\,;\,FM=PN$ .

Tam giác EFM có $\widehat{E}={{90}^{o}}$ $\Rightarrow \,\Delta EFM$ vuông tại E $\Rightarrow \,{{S}_{EFM}}=\dfrac{1}{2}EF.EM$ .

Tam giác KPN có $\widehat{K}={{90}^{o}}$ $\Rightarrow \,\Delta KPN$ vuông tại K $\Rightarrow \,{{S}_{KPN}}=\dfrac{1}{2}KP.KN$ .

Khi đó: $\dfrac{{{S}_{EFM}}}{{{S}_{KPN}}}=\dfrac{EF.EM}{KP.KN}=\dfrac{KP.KN}{KP.KN}=1.$

Vì $\Delta ABC=\Delta KPN$ nên $AB=KP=9cm\,;$ $AC=KN=14cm\,;\,BC=PN.$

Vì chu vi tam giác KPN bằng 30cm nên $KP+KN+PN=30\,(cm)$

$\Rightarrow \,PN=30-(KP+KN)=30-(9+14)=7\,(cm).$

Vậy $BC=PN=7cm.$

$\Delta ABC=\Delta DEF\,\Rightarrow \,\,\widehat{A}=\widehat{D}\,;\,\widehat{B}=\widehat{E}={{55}^{o}}\,;\,\widehat{C}=\widehat{F}={{60}^{o}}$ .

Xét tam giác DEF có $\widehat{D}+\widehat{E}+\widehat{F}={{180}^{o}}$

$\Rightarrow \,\,\widehat{D}={{180}^{o}}-\left( \widehat{E}+\widehat{F} \right)={{180}^{o}}-\left( {{55}^{o}}+{{60}^{o}} \right)={{65}^{o}}$ .

Vì $\Delta ABC=\Delta MNP$ $\Rightarrow \,\,\widehat{A}=\widehat{M}\,;\,\widehat{B}=\widehat{N}\,={{70}^{o}};\,\widehat{C}=\widehat{P}={{40}^{o}}$ .

Xét tam giác MNP có: $\widehat{M}+\widehat{N}+\widehat{P}={{180}^{o}}$

$\Rightarrow \,\,\widehat{M}={{180}^{o}}-\left( \widehat{N}+\widehat{P} \right)={{180}^{o}}-\left( {{70}^{o}}+{{40}^{o}} \right)={{70}^{o}}$

Vậy tam giác MNP có $\widehat{M}=\widehat{N}={{70}^{o}}$ .

Ta có $\widehat{A}+\widehat{B}+\widehat{C}={{180}^{o}}$ mà $\widehat{A}=\widehat{B}=2\widehat{C}$ nên $\left\{ \begin{array}{l} \widehat{A}={{72}^{o}} \\ \widehat{B}={{72}^{o}} \\ \widehat{C}={{36}^{o}} \end{array} \right.$

Mà $\Delta ABC=\Delta DEG\Rightarrow \widehat{E}=\widehat{B}={{72}^{o}}$

Do $\Delta ABC=\Delta MNP\Rightarrow \left\{ \begin{array}{l} AB=MN \\ BC=NP \\ AC=MP \end{array} \right.$

Chu vi tam giác $MNP$ là $MN+NP+MP=MN+BC+AC=6+9+8=23\,cm$

Vì $AB=QP\,;\,BC=E\,Q$ và $\Delta ABC$ bằng một tam giác có ba đỉnh P, E, Q nên AC = PE.

Khi đó hai tam giác bằng nhau theo thứ tự đỉnh tương ứng là: $\Delta ABC=\Delta PQE$ .

Ta có $\Delta PQ\text{R}=\Delta SIK\Rightarrow \left\{ \begin{array}{l} \widehat{P}=\widehat{S} \\ \widehat{Q}=\widehat{I} \\ \widehat{R}=\widehat{K} \end{array} \right.$

Mặt khác $\widehat{P}+\widehat{Q}+\widehat{R}=\widehat{P}+\widehat{I}+\widehat{R}={{180}^{o}}\Rightarrow \widehat{R}={{40}^{0}}$

$\Delta ABC=\Delta MNP$

$\Rightarrow \,\widehat{A}=\widehat{M}\,;\,\widehat{B}=\widehat{N}\,={{50}^{o}}\,;\,\widehat{C}=\widehat{P}={{85}^{o}}$

Xét tam giác ABC có: $\widehat{A}+\widehat{B}+\widehat{C}={{180}^{o}}$

$\Rightarrow \,\widehat{A}={{180}^{o}}-\left( \widehat{B}+\widehat{C} \right)={{180}^{o}}-({{50}^{o}}+{{85}^{o}})={{45}^{o}}$ .

$\Rightarrow \,\widehat{M}=\widehat{A}={{45}^{o}}.$

$\Delta ABC=\Delta KPN$ $\Rightarrow \,AB=KP\,;\,AC=KN\,;\,BC=PN$ .

Chu vi tam giác ABC là: ${{C}_{ABC}}=AB+AC+BC$ .

Chu vi tam giác KPN là: ${{C}_{KPN}}=KP+KN+PN$ .

Do đó: $\dfrac{{{C}_{ABC}}}{{{C}_{KPN}}}=\dfrac{AB+AC+BC}{KP+KN+PN}=1.$

Ta có $\Delta ABC=\Delta DEF\Rightarrow \left\{ \begin{array}{l} AC=DF \\ AB=DE \\ BC=\text{EF} \end{array} \right.$

Chu vi tam giác $DEF$ là $DF+DE+F\text{E}=DF+AB+BC=6+5+7=18\,\,cm$

Vì $\Delta ABC=\Delta MNP$ $\Rightarrow \,\,\widehat{A}=\widehat{M}={{60}^{o}}\,;\,\widehat{B}=\widehat{N}\,;\,\widehat{C}=\widehat{P}={{30}^{o}}$ .

Xét tam giác ABC có: $\widehat{A}+\widehat{B}+\widehat{C}={{180}^{o}}$

$\Rightarrow \,\,\widehat{B}={{180}^{o}}-\left( \widehat{A}+\widehat{C} \right)={{180}^{o}}-\left( {{60}^{o}}+{{30}^{o}} \right)={{90}^{o}}$

$\Rightarrow \,\Delta ABC$ vuông tại B.