# Calculate the sum of the line segments DA and DC?

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.

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What does PQRS have to do with anything? Where is D in relation to ABC? – anon Apr 1 '12 at 8:44
@robjohn i am sorry.edited the question. – vikiiii Apr 1 '12 at 8:46
Hint: Ptolemy's theorem. – dtldarek Apr 1 '12 at 8:56

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

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