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Is a quadrilateral with sides lengths $3$, $3$, $4$, and $4$ cyclic?

Progress

I found that sides joining 3 and 4 are of equal length. then I found that other diagonal should also have same length as the first one, then couldn't find the length and got stuck.

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The Pragmatick's answer is (currently) incorrect. Specifically, $AD$ (or $AC$) does not need to be the diameter.

Unlike in triangles, determining the side lengths of a quadrilateral does not guarantee a unique quadrilateral. The angle between $AB$ and $AC$ can vary, giving quadrilaterals that aren't cyclic.

If, however, you are wondering whether or not this quadirlateral is tangential, then The Pragmatick's first line is helpful. From the congruency we can tell the angle bisectors of the angle concur, so this would have to be the 'incenter'. More generally, in a quadrilateral if $AB + CD = AC + BD$, the quadrilateral is tangential.

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A quadrilateral with sides $3,3,4,4$ is cyclic $\iff AC=5$. Consider this figure:

enter image description here

Here, $AD = AB = 3$. $DC = CB = 4$. Therefore $\triangle ADC \cong\triangle ABC$

$\implies AC$ must be the diameter. If $AC$ is the diameter, angles subtended on circle must be $90^\circ$ $(\angle ADC=\angle ABC=90)$

$\therefore AC=\sqrt{3^2+4^2}=5$

EDIT

@cyclicduck Yes you are correct. I didn't read the question properly. The quadrilateral $ABCD$ will be cyclic $\iff AC$ is the diameter, or $AC=5$

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  • $\begingroup$ how can i find whether a circle can be inscribed in the particular quadrilateral. $\endgroup$ – me0001 Dec 15 '14 at 14:18
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    $\begingroup$ @me0001 You can coordinate-bash. ;) $\endgroup$ – Ahaan S. Rungta Dec 15 '14 at 15:09
  • $\begingroup$ See if the perpendicular bisectors from AB and BC meet on AC and their lengths are the same - that would be the center of an inscribed circle. $\endgroup$ – marty cohen Dec 16 '14 at 3:44
  • $\begingroup$ @me0001 Take cyclicduck's answer, its correct .. :) $\endgroup$ – Sawarnik Dec 16 '14 at 6:41

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