$BE=x, FC=y, BC=a$
Then prove that $x^{2/3}+ y^{2/3}= a^{2/3}$ 
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hmakholm left over Monica
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user9640947
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2Try to apply your decisions and in the future use LaTeX – Vladislav Kharlamov May 12 '18 at 09:48
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Why down votes??? – user9640947 May 12 '18 at 09:51
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Poor design, lack of ideas. – Vladislav Kharlamov May 12 '18 at 09:53
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This is not to mention the $ \frac{x}{3}+\frac{y}{3} = \frac{b}{3} \Leftrightarrow x+y = b$ – Vladislav Kharlamov May 12 '18 at 09:54
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I think it is all clear. Design and data are sufficient to understand the problem – user9640947 May 12 '18 at 09:55
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All is clear, but the design is still not very, here it is accepted to respect other members, so that it would be pleasant for them to look at your question – Vladislav Kharlamov May 12 '18 at 09:56
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I can try to fix the design, but you will have to attach the ideas yourself – Vladislav Kharlamov May 12 '18 at 09:57
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x^2/3+y^2/3=a^2/3 is the same as $x^2/3+y^2/3=a^2/3 $, which is the same as $x^2+y^2=a^2$ surely? – Angina Seng May 12 '18 at 09:59
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Brother mathematics always made to test imagination and approach.if all things become simple then all level of become equal. It will become simple when one solve it in solution with nice idea ok ?? – user9640947 May 12 '18 at 10:01
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There's planty of similar triangles there to start working with. Don't you have any progress to share with us? Or do you simply want someone to do your homework for you? – hmakholm left over Monica May 12 '18 at 10:16
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I think it need additional construction of circle in which power of point theorem may be use – user9640947 May 12 '18 at 10:19
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Basic trigonometry on the many similar right triangles, and some algebra will be enough. Try to express each of $x$ and $y$ in terms of $a$ and the angle at $C$. – hmakholm left over Monica May 12 '18 at 10:25
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@HenningMakholm brother then use that and please sove this. – user9640947 May 12 '18 at 10:27
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@user9640947: I'm not your brother. And it's your homework. We're not a homework solving service. – hmakholm left over Monica May 12 '18 at 10:27
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@HenningMakholm ok human being?? If you don't know how solve it.then dont make excuses.its not place for debate.and it my homework or what so – user9640947 May 12 '18 at 10:29
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2If Henning says he can solve it, then Henning can solve it. Henning has given you some ideas on how to do it – why not try to follow Henning's hint? – Gerry Myerson May 12 '18 at 10:44
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You all are in team to just downvote nothing else. – user9640947 May 12 '18 at 10:49
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$$x = |BD|\cos B = |AB|\cos^2 B= a \cos^3 B \qquad\qquad y = a \sin^3 B$$ – Blue May 13 '18 at 03:14
1 Answers
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By similar triangles $\triangle CDF,\triangle ABC$ $$\dfrac{CF}{BC}=\dfrac{AE}{AB}\implies\dfrac y{a\sin B}=\dfrac{a\cos B-x}{a\cos B} \iff x\sin B+y\cos B=a\cos B\sin B\ \ \ \ (1)$$
Similarly, by similar triangles $\triangle DAF,\triangle ABC$
$$\dfrac{AF}{CA}=\dfrac{DF}{AB}\implies\dfrac{a\cos B-x}{a\sin B}=\dfrac{a\sin B-y}{a\cos B}\iff x\cos B-y\sin B=a\cos2B\ \ \ \ (2)$$
Solve $(1),(2)$ for $x,y$ to find $x=a\cos^3B,y=a\sin^3B$
Use Prove $\sin^2\theta + \cos^2\theta = 1$
Alternatively, let $AB=c,CA=b\implies a^2=b^2+c^2\ \ \ \ (3)$
by similarities of the triangles we have $$\dfrac y{c-x}=\dfrac bc\ \ \ \ (4)$$ and $$\dfrac{c-x}{b-y}=\dfrac bc\ \ \ \ (5)$$
Solve $(4),(5)$ for $x,y$ and use $(3)$
lab bhattacharjee
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