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Question Number 194056 by ajfour last updated on 26/Jun/23

Answered by a.lgnaoui last updated on 27/Jun/23

$$\mathrm{ABC}\mathrm{triangle}\mathrm{equilaterale}$$$$\mathrm{AB}=\mathrm{BC}=\mathrm{AC}\mathrm{BC}=2\mathrm{acos}30=\mathrm{a}\sqrt{3}$$$$\Rightarrow \mathrm{AH}=\mathrm{BCsin}60=\frac{3\mathrm{a}}{2}$$$$\mathrm{AI}=\mathrm{AM}+\mathrm{MI}=2\mathrm{a}\left(\mathrm{i}\right)$$$$\measuredangle \mathrm{MAN}=30;\sin 30=\frac{\mathrm{c}}{\mathrm{AM}}\Rightarrow \mathrm{AM}=2\mathbf{c}$$$$\mathrm{MI}=\mathrm{c}+2\mathrm{b}$$$$\left(\mathrm{i}\right)\Rightarrow 2\mathbf{a}=3\mathbf{c}+2\mathbf{b}$$$$$$$$\mathbf{c}=\frac{2(\mathbf{a}-\mathbf{b})}{3}$$$$$$

Commented by ajfour last updated on 27/Jun/23

$$x+3=y$$$$\Rightarrow x=0because$$$$yshouldbe3.$$

Commented by a.lgnaoui last updated on 27/Jun/23

Commented by mr W last updated on 27/Jun/23

$$triangleisnotequilateral!$$

Answered by mr W last updated on 27/Jun/23

$$\frac{c}{b}=\frac{2a-2b-c}{2a-b}$$$$\Rightarrow c=\frac{(a-b)b}{a}$$

Commented by ajfour last updated on 30/Jun/23

$$Excellent$$

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