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ABCD is a cyclic quadrilateral and the points A , B and C form an equilateral triangle .

Then what is the sum of the length of line segments DA and DC? What property should i use to get the value of DA and DC.

I need a hint to start the solution.

Thanks in advance.

  • 0
    Hint: [Ptolemy's theorem](http://en.wikipedia.org/wiki/Ptolemy%27s_theorem).2012-04-01

3 Answers 3

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Let's denote :

$|AB|=|BC|=|AC|=a$

$|CD|=b$

$|AD|=c$

Since quadrilateral is cyclic and $\Delta ABC$ is equilateral it follows that :

$\frac{1}{2}(a^2+bc)\cdot \sin 60^{\circ} =\frac{a^2 \sqrt 3}{4}+\sqrt{s(s-a)(s-b)(s-c)}$

where $s=\frac{a+b+c}{2}$

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The given information does not suffice to compute $|DA|+|DC|$, as $D$ can be any point on the shorter arc between $A$ and $C$. All you can say is the following: In the triangle $ADC$ the angle at $D$ is $120^\circ$, whence by the cosine theorem on has $|DA|^2+|DC|^2 + |DA|\ |DC|=|AC|^2=3 r^2\ ,$ where $r$ is the radius of the circumscribing circle.

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Use Ptolemy's theorem,

Let AB = BC = AC = a Then by Ptolemy's theorem, AC x BD = AB x CD + BC x AD a x BD = a ( DA + DC ) BD = DA + DC