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Let the sides $AB,BC,CD,DA$ of a cyclic quadrilateral $ABCD$ are in Geometric Progression with common ratio $2$.then the question is to evaluate $\frac{BD}{AB}$.

I tried using Ptolemy Theorem and took the sides as $(\frac{a}{2√2},\frac{a}{√2},a√2,2a√2)$.From this I could find $BD \times AC$ in terms of $a$ but am facing trouble evaluating $AC \times AB$ .Any hint would be highly appreciated.Thanks.

Edit-- As the answer below points out no quadrilateral is possible if common ratio 2. What if it is $r$ and given that such a quadrilateral exists.

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No such quadrilateral exists, whether cyclic or not.

Assume WLOG the shortest side is $1\,$, then the others must be $2,4,8\,$. By the triangle inequality, any side of a polygon can be no larger than the sum of the others. But $8 \gt 1+2+4\,$.


[ EDIT ] For ratios $r$ that allow such a polygon to be constructed, let $AB=a\,$, $BC=r\,a\,$, $CD=r^2\,a\,$, $DA=r^3\,a\,$. By homogeneity the ratio $BD/AB$ will be the value of $BD$ when $a=1$.

Let $\alpha = \angle BAD\,$, then $\angle BCD = \pi - \alpha$ since $ABCD$ is cyclic. Express $BD$ (for $a=1$) using the law of cosines twice in $\triangle BAD\,$, $\triangle CBD\,$ respectively:

$$ BD^2 = 1 + r^6 - 2 r^3 \cos \alpha = r^2 + r^4 + 2 r^3 \cos \alpha $$

The latter equality gives: $$\cos \alpha = \cfrac{1}{4 r^3}(r^6-r^4-r^2+1)\,$$

Substituting back into either of the previous: $$BD^2 = \cfrac{1}{2}(r^6+r^4+r^2+1)$$

So in the end:

$$\cfrac{BD}{AB} = \sqrt{\cfrac{r^6+r^4+r^2+1}{2}}$$

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  • $\begingroup$ What if it the common ratio is r such that the quadrilateral is possible. $\endgroup$ – user408949 Jan 22 '17 at 6:57

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