# Prove that $x^{2/3}+ y^{2/3}= a^{2/3}$ [closed]

$BE=x, FC=y, BC=a$
Then prove that $x^{2/3}+ y^{2/3}= a^{2/3}$

• Try to apply your decisions and in the future use LaTeX – Vladislav Kharlamov May 12 '18 at 9:48
• Why down votes??? – user9640947 May 12 '18 at 9:51
• Poor design, lack of ideas. – Vladislav Kharlamov May 12 '18 at 9:53
• 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 9:54
• If 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

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 $x=a\cos^3B,y=a\sin^3B$

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 $(3)$