<s>[INST] How to show a $AE=\frac{2+\sqrt{7}}{3}$ I have been thinking about this problem for quite some time and unable to find any clue. I also have some troubles uploading an image here, but hopefully the question is clear enough.
Let $$\angle B=\dfrac{\pi}{3},DE\perp AC,DE=\dfrac{\sqrt{3}}{2},CD=\dfrac{\sqrt{7}}{2},AB+BC+AC=6$$
show that
$$AE=\dfrac{2+\sqrt{7}}{3}$$

I did $EC=1$,$AE=x$,then $AD=\sqrt{x^2+\frac{3}{4}}$,I am trying to use the property by using law of cosines in triangles $BDC$ and $ABC$
I really appreciate any helps! Many thanks!
 [/INST] Call $AE=x$. As you noticed $EC=1$ and $AD=\sqrt { x^2 +3/4}$. Call $\angle BAC= \theta$. We have
$$ \sin \theta = \frac{DE}{AD}=\frac{\sqrt 3}{\sqrt{4x^2+3}}.$$
By the law of sines in $ABC$
$$\frac{AC}{\sin 60}=\frac{BC}{\sin \theta}$$
and we get 
$$BC=\frac{2(x+1)}{\sqrt{4x^2+3}}. $$
Now we compute $\cos \theta=\frac{2x}{\sqrt{4x^2+3}}$ and
$$\sin \angle ACB =\sin (60+\theta)=\frac{x\sqrt 3}{\sqrt{4x^2+3}}+\frac{\sqrt 3}{2\sqrt{4x^2+3}}=\frac{(2x+1)\sqrt 3}{2\sqrt{4x^2+3}}.$$
Applying again the law of sines we get
$$AB=\frac{(2x+1)(x+1)}{\sqrt{4x^2+3}}.$$ 
Finally we are able to solve for $x$ using $AB+BC+CA=6$:
$$x+1+\frac{(2x+1)(x+1)}{\sqrt{4x^2+3}}+\frac{2(x+1)}{\sqrt{4x^2+3}}=6 $$
that simplifies (after some work) to
$$ (10x-11)(x^2+1)=0$$
hence $AE=x=\frac{11}{10}$.
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