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I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?

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Yes, it's true.

Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.

Also, let $AE$ and $BE$ be bisectors of $\angle BAD$ and $\angle ABC$ respectively.

Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.

Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.

Thus, $$AB+CD_1=BC+AD_1$$ and $$AB+CD=BC+AD$$ from the given.

Hence, $$CD_1-CD=AD_1-AD$$ or $$CD_1-CD=\pm DD_1,$$ which by the triangle inequality is possible, when $D\equiv D_1$ and we are done!

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  • $\begingroup$ Thank you very much. The proof was great! $\endgroup$ – Shashwat1337 Dec 30 '18 at 17:37
  • $\begingroup$ @Shashwat1337 You are welcome! $\endgroup$ – Michael Rozenberg Dec 30 '18 at 17:38

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