# Does the equality of the sum of opposite sides in a quadrilateral necessarily imply the existence of an inscribed circle?

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?

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!

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