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Question Number 6890 by Tawakalitu. last updated on 01/Aug/16

Commented by Yozzii last updated on 02/Aug/16

Let EC=x. BD=3, DA=5  BD=BE since BD and BE are lines  tangent to the same circle from   the common point B.  ∴ BE=3. Similarly, DA=AF=5,  and EC=FC=x>0, since the   inscribed circle touches all sides of  △ABC.  By the law of cosines  BC^2 =AB^2 +AC^2 −2(AB)(AC)cos∠BAC  BC=BE+EC=3+x  AB=BD+DA=3+5=8  AC=AF+FC=5+x  (3+x)^2 =(3+5)^2 +(5+x)^2 −2(3+5)(5+x)cos60°  x^2 +6x+9=64+x^2 +10x+25−8(5+x)  6x=55+25+10x−40−8x  4x=40  x=10  ⇒BC=3+10=13

$${Let}\:{EC}={x}.\:{BD}=\mathrm{3},\:{DA}=\mathrm{5} \\ $$$${BD}={BE}\:{since}\:{BD}\:{and}\:{BE}\:{are}\:{lines} \\ $$$${tangent}\:{to}\:{the}\:{same}\:{circle}\:{from}\: \\ $$$${the}\:{common}\:{point}\:{B}. \\ $$$$\therefore\:{BE}=\mathrm{3}.\:{Similarly},\:{DA}={AF}=\mathrm{5}, \\ $$$${and}\:{EC}={FC}={x}>\mathrm{0},\:{since}\:{the}\: \\ $$$${inscribed}\:{circle}\:{touches}\:{all}\:{sides}\:{of} \\ $$$$\bigtriangleup{ABC}. \\ $$$${By}\:{the}\:{law}\:{of}\:{cosines} \\ $$$${BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\mathrm{2}\left({AB}\right)\left({AC}\right){cos}\angle{BAC} \\ $$$${BC}={BE}+{EC}=\mathrm{3}+{x} \\ $$$${AB}={BD}+{DA}=\mathrm{3}+\mathrm{5}=\mathrm{8} \\ $$$${AC}={AF}+{FC}=\mathrm{5}+{x} \\ $$$$\left(\mathrm{3}+{x}\right)^{\mathrm{2}} =\left(\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}+{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}+\mathrm{5}\right)\left(\mathrm{5}+{x}\right){cos}\mathrm{60}° \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}=\mathrm{64}+{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{25}−\mathrm{8}\left(\mathrm{5}+{x}\right) \\ $$$$\mathrm{6}{x}=\mathrm{55}+\mathrm{25}+\mathrm{10}{x}−\mathrm{40}−\mathrm{8}{x} \\ $$$$\mathrm{4}{x}=\mathrm{40} \\ $$$${x}=\mathrm{10} \\ $$$$\Rightarrow{BC}=\mathrm{3}+\mathrm{10}=\mathrm{13} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 02/Aug/16

Wow, i really appreciate. Thanks so much.

$${Wow},\:{i}\:{really}\:{appreciate}.\:{Thanks}\:{so}\:{much}. \\ $$

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