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

a) Xét ΔAEB vuông tại E và ΔAFC vuông tại F có 

\(\widehat{FAC}\) chung

Do đó: ΔAEB\(\sim\)ΔAFC(g-g)

Suy ra: \(\dfrac{AE}{AF}=\dfrac{AB}{AC}\)(Các cặp cạnh tương ứng tỉ lệ)

hay \(AE\cdot AC=AF\cdot AB\)(ĐPCM)

b)

Ta có: \(\dfrac{AE}{AF}=\dfrac{AB}{AC}\)(cmt)

nên \(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét ΔAEF và ΔABC có 

\(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)(cmt)

\(\widehat{FAE}\) chung

Do đó: ΔAEF\(\sim\)ΔABC(c-g-c)

Ta có:  B O C ^ = 2 B A C ^ ,   C O A ^ = 2 C B A ^ ,   A O B ^ = 2 A C B ^

( góc ở tâm gấp 2 lần số đo góc nội tiếp cùng chắn 1 cung )

S = S O A B + S O B C + S O C A  

=  1 2 O A . O B . sin A O B ^ + 1 2 O B . O C . sin B O C ^ + 1 2 O C . O A . sin C O A ^

S = 1 2 R 2 sin 2 A + sin 2 B + sin 2 C  .

ĐÁP ÁN A

mình cần gâps huhu

=>AM=AN

Vì M nằm trong tam giác ABC nên ta có:

Khi đó điểm M nhìn các cạnh AB,BC,CA của tam giác ABC dưới một góc bằng 120 °

Ta có thẻ dựng điểm M như sau:

Dựng cung chứa góc  120 °  vẽ trên đoạn BC

Dựng cung chứa góc  120 °  vẽ trên đoạn AC

Giao điểm thứ hai ngoài C của hai cung này là điểm M cần dựng

a)

Nhận xét: H là một điểm nằm trong tam giác ABC.

b)

Nhận xét: H trùng với đỉnh A của tam giác ABC.

c)

Nhận xét: H nằm ngoài tam giác ABC.

a) Tính các góc của tam giác

Ta có: $BC^2 = 14^2 = 196$

$AB^2 + AC^2 = 4,5^2 + 13^2 = 20,25 + 169 = 189,25$

Vì: $BC^2 > AB^2 + AC^2$ nên $\widehat{A}$ là góc tù.

Áp dụng định lý cosin:

$\cos A = \dfrac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}$

$= \dfrac{20,25 + 169 - 196}{2 \cdot 4,5 \cdot 13}$

$= \dfrac{-6,75}{117}$

$\approx -0,0577$

Suy ra: $\widehat{A} \approx 93^\circ$

Áp dụng định lý cosin với góc $B$:

$\cos B = \dfrac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$

$= \dfrac{20,25 + 196 -169}{2 \cdot 4,5 \cdot 14}$

$= \dfrac{47,25}{126}$

$\approx 0,375$

=> $\widehat{B} \approx 68^\circ$

Góc còn lại: $\widehat{C} = 180^\circ - 93^\circ - 68^\circ \approx 19^\circ$

Vậy: $\widehat{A} \approx 93^\circ,\ \widehat{B} \approx 68^\circ,\ \widehat{C} \approx 19^\circ$

b) Tính diện tích tam giác $ABC$

Nửa chu vi là: $p = \dfrac{4,5 + 14 + 13}{2} = 15,75$

Diện tích tam giác: $S = \sqrt{p(p-AB)(p-BC)(p-AC)}$

$= \sqrt{15,75(15,75-4,5)(15,75-14)(15,75-13)}$

$= \sqrt{15,75 \cdot 11,25 \cdot 1,75 \cdot 2,75}$

$\approx \sqrt{852,54}$

$\approx 29,2$

Vậy diện tích tam giác $ABC$ xấp xỉ: $29,2\ \text{đơn vị diện tích}$.