Question

[Linh Nguyễn]

Tứ giác ABCD có AB = AD, \(\widehat{BAD}=60^o,\widehat{BCD}=120^o\). Chứng minh CA = CB + CD

Answer 1

[Pham Van Hung]

C B A D I

Answer 2

[Pham Van Hung]

Trên tia đối của DC lấy I sao cho DI = CB

Khi đó: \(CB+CD=DI+CD=IC\)

Tứ giác ABCD có: \(\widehat{BAD}+\widehat{BCD}=60^0+120^0=180^0\)

\(\Rightarrow\widehat{ADC}+\widehat{ABC}=180^0\)

Mà \(\widehat{ADC}+\widehat{ADI}=180^0\Rightarrow\widehat{ABC}=\widehat{ADI}\)

\(\Delta BAD:AB=AD,\widehat{BAD}=60^0\Rightarrow\Delta BAD\) đều 

\(\Rightarrow\widehat{BAD}=60^0\)

\(\Delta ABC=\Delta ADI\left(c.g.c\right)\Rightarrow\hept{\begin{cases}\widehat{BAC}=\widehat{DAI}\\AC=AI\end{cases}}\)

\(\widehat{CAI}=\widehat{CAD}+\widehat{DAI}=\widehat{CAD}+\widehat{BAC}=\widehat{BAD}=60^0\)

Tam giác ACI đều nên AC = AI = CI

Mà \(CB+CD=IC\Rightarrow CA=CB+CD\)