Question

[Phan Thanh Tịnh]

Lấy điểm D trong \(\Delta ABC\)cân tại A sao cho AD = BC.Biết\(\widehat{BAD}+3\widehat{CAD}=60^0\) ,tính \(\widehat{DCA}\)theo \(\widehat{BAD}\)và\(\widehat{CAD}\)

Answer 1

[Nguyễn Tất Đạt]

Trên nửa mặt phẳng bờ AD, dựng tam giác đều ADE khác phía với điểm C. Nối E với C.

\(\Delta\)ADE đều => AD=ED=AE và ^DAE=^DEA=ADE=60⁰.

Có: AD=BC => AE=BC

Ta có: ^EAC=^DAE+^CAD=\(60^0+\widehat{CAD}\) \(\left(1\right)\)

Xét \(\Delta\)ABC: Cân tại A => ^B=^C= \(\frac{180^0-\widehat{BAC}}{2}=\frac{120^0+60^0-\widehat{BAC}}{2}\)

Thay \(\widehat{BAD}+3\widehat{CAD}=60^0\) và \(\widehat{BAD}+\widehat{CAD}=\widehat{BAC}\) vào biểu thức trên, ta được:

\(\widehat{ABC}=A\widehat{CB}=\frac{120^0+\widehat{BAD}+3\widehat{CAD}-\left(\widehat{BAD}+\widehat{CAD}\right)}{2}\)

\(=\frac{120^0+2\widehat{CAD}}{2}=\frac{2\left(60^0+\widehat{CAD}\right)}{2}=60^0+\widehat{CAD}\)\(\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\widehat{EAC}=\widehat{ACB}=60^0+\widehat{CAD}\)

Xét \(\Delta\)ABC và \(\Delta\)CEA có:

BC=EA

^ACB=^EAC         \(\Rightarrow\Delta ABC=\Delta CEA\left(c.g.c\right)\)

AC chung

\(\Rightarrow AB=CE\)(2 cạnh tương ứng). Mà \(AB=AC\Rightarrow AC=CE\)

Xét \(\Delta\)ADC và \(\Delta\)EDC có:

AD=ED

DC chung        \(\Rightarrow\Delta ADC=\Delta EDC\left(c.c.c\right)\)

AC=EC

\(\Rightarrow\widehat{ACD}=\widehat{ECD}=\frac{1}{2}\widehat{ECA}\)(2 góc tương ứng). Mà \(\Delta ABC=\Delta CEA\)(cmt)

\(\Rightarrow\widehat{BAC}=\widehat{ECA}\)(2 góc tương ứng) \(\Rightarrow\widehat{ACD}=\frac{1}{2}\widehat{BAC}=\frac{1}{2}\left(\widehat{BAD}+\widehat{CAD}\right)=\frac{\widehat{BAD}+\widehat{CAD}}{2}\)

Hay \(\widehat{DCA}=\frac{\widehat{BAD}+\widehat{CAD}}{2}\).

Answer 2

[Nguyễn Tất Đạt]

Còn 3 cách nữa ! :v

* Cách 2:

Trên nửa mặt phẳng bờ BC chứa điểm A, dựng \(\Delta\)BCF đều.

=> BF=CF=BC và ^BFC=^FBC=^FCB=60⁰.

AD=BC => AD=CF.

Ta có: \(\widehat{ABC}=\widehat{ACB}=\frac{180^0-\widehat{BAC}}{2}=\frac{3.60^0-\left(\widehat{BAD}+\widehat{CAD}\right)}{2}\)

\(=\frac{3.\left(\widehat{BAD}+3\widehat{CAD}\right)-\left(\widehat{BAD}+\widehat{CAD}\right)}{2}=\frac{3\widehat{BAD}+9\widehat{CAD}-\left(\widehat{BAD}+\widehat{CAD}\right)}{2}\)

\(=\frac{2\widehat{BAD}+8\widehat{CAD}}{2}=\frac{2\left(\widehat{BAD}+4\widehat{CAD}\right)}{2}=\widehat{BAD}+4\widehat{CAD}\)

Ta có: \(\widehat{FCA}=\widehat{ACB}-\widehat{FCB}=\widehat{ACB}-60^0\)

Thay \(\widehat{ACB}=\widehat{BAD}+4\widehat{CAD}\)và \(\widehat{BAD}+3\widehat{CAD}=60^0\)vào biểu thức trên ta có:

\(\widehat{FCA}=\widehat{BAD}+4\widehat{CAD}-\left(\widehat{BAD}+3\widehat{CAD}\right)=\widehat{CAD}\)\(\Rightarrow\widehat{FCA}=\widehat{CAD}\)

\(\Rightarrow\Delta FAC=\Delta DCA\left(c.g.c\right)\Rightarrow\widehat{FAC}=\widehat{DCA}\)(2 góc tương ứng)

Mà \(\Delta FAB=\Delta FAC\left(c.c.c\right)\Rightarrow\widehat{FAB}=\widehat{FAC}=\frac{\widehat{BAC}}{2}\)

\(\Rightarrow\widehat{FAC}=\widehat{DCA}=\frac{\widehat{BAC}}{2}\Rightarrow\widehat{DCA}=\frac{\widehat{BAD}+\widehat{CAD}}{2}.\)

* Cách 3:

Trên nửa mặt phẳng bờ AB có chứa điểm C, dựng \(\Delta ABI\)đều.

\(\Rightarrow AB=BI=AI\)và \(\widehat{BAI}=\widehat{ABI}=\widehat{AIB}=60^0\)

Mà \(AB=AC\Rightarrow AC=BI\).

Ta có: \(\widehat{CBI}=\widehat{ABC}-\widehat{ABI}=\frac{180^0-\widehat{BAC}}{2}-60^0=\widehat{CAD}\)(C/m tương tự cách 2)

\(\Rightarrow\Delta BCI=\Delta ADC\left(c.g.c\right)\Rightarrow\widehat{CIB}=\widehat{DCA}\)(2 góc tương ứng)

Lại có: \(\widehat{CAI}=\widehat{BAI}-\widehat{BAC}=60^0-\widehat{BAC}=\widehat{BAD}+3\widehat{CAD}-\left(\widehat{BAD}+\widehat{CAD}\right)\)

\(\Leftrightarrow\widehat{CAI}=2\widehat{CAD}\).

\(AC=AB=AI\Rightarrow\Delta CAI\)cân tại A \(\Rightarrow\widehat{ACI}=\widehat{AIC}=\frac{180^0-\widehat{CAI}}{2}=\frac{3.60^0-2\widehat{CAD}}{2}\)

\(\Leftrightarrow\widehat{AIC}=\frac{3.\left(\widehat{BAD}+3\widehat{CAD}\right)-2\widehat{CAD}}{2}=\frac{3\widehat{BAD}+9\widehat{CAD}-2\widehat{CAD}}{2}\)

\(\Leftrightarrow\widehat{AIC}=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}\)

Nhận thấy:

 \(\widehat{CIB}=\widehat{AIC}-\widehat{AIB}=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-60^0=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-\left(\widehat{BAD}+3\widehat{CAD}\right)\)

\(=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-\frac{2\widehat{BAD}+6\widehat{CAD}}{2}=\frac{\widehat{BAD}+\widehat{CAD}}{2}\)

\(\Rightarrow\widehat{CIB}=\frac{\widehat{BAD}+\widehat{CAD}}{2}\). Mà \(\widehat{CIB}=\widehat{DCA}\)(cmt) \(\Rightarrow\widehat{DCA}=\frac{\widehat{BAD}+\widehat{CAD}}{2}.\)

* Cách 4: 

Trên nửa mặt phẳng bờ AC không chứa điểm B, dựng \(\Delta ACK\)đều.

\(\Rightarrow AC=AK=CK\)và \(\widehat{CAK}=\widehat{ACK}=\widehat{AKC}=60^0\).

Ta có: \(\widehat{DAK}=\widehat{CAD}+\widehat{CAK}=\widehat{CAD}+60^0=\widehat{ABC}\)(c/m tương tự cách 1 ở câu trả lời trước)

\(\Rightarrow\Delta AKD=\Delta BAC\left(c.g.c\right)\)\(\Rightarrow\widehat{BAC}=\widehat{AKD}\)(2 góc tương ứng)

\(\Rightarrow\widehat{AKD}=\widehat{BAD}+\widehat{CAD}\).

\(AC=KD\)( 2 cạnh tương ứng) \(\Rightarrow KD=KC\Rightarrow\Delta DKC\)cân tại K 

Lại có: \(\widehat{DKC}=\widehat{AKC}-\widehat{AKD}=60^0-\left(\widehat{BAD}+\widehat{CAD}\right)\)

\(=\widehat{BAD}+3\widehat{CAD}-\left(\widehat{BAD}+\widehat{CAD}\right)=2\widehat{CAD}\)\(\Rightarrow\widehat{DKC}=2\widehat{CAD}\)

\(\Delta DKC\)cân tại K (cmt) \(\Rightarrow\widehat{KDC}=\widehat{KCD}=\frac{180^0-\widehat{DKC}}{2}=\frac{3.60^0-2\widehat{CAD}}{2}\)

\(=\frac{3\widehat{BAD}+9\widehat{CAD}-2\widehat{CAD}}{2}=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}\)

\(\widehat{DCA}=\widehat{KCD}-\widehat{ACK}=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-60^0=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-\left(\widehat{BAD}+3\widehat{CAD}\right)\)

\(\Rightarrow\widehat{DCA}=\frac{3\widehat{BAD}+7\widehat{CAD}}{2}-\frac{2\widehat{BAD}+6\widehat{CAD}}{2}=\frac{\widehat{BAD}+\widehat{CAD}}{2}\)

\(\Rightarrow\widehat{DCA}=\frac{\widehat{BAD}+\widehat{CAD}}{2}.\)