Prove that the bisectors of angles $B A D, B C D$ and $B E C$ intersect at one point.

## Solution

Necessity. Given: $A B C D$ is a circumscribed quadrilateral. Let the tangents from points $A, B, C, D, E$ and $F$ to the inscribed circle be $a, b, c, d, e$ and $f$ respectively. Then,

$$E A=e-a, A F=a+f, E C=e+c, C F=f-c$$

Thus,

$$E A+A F=(e-a)+(a+f)=e+f, E C+C F=(e+c)+(f-c)=e+f$$

Therefore, $E A+A F=E C+C F$.

Sufficiency. Suppose the equality $E A+A F=E C+C F$ holds. We need to prove that the bisectors of angles $B A D, B C D$ and $B E C$ intersect at one point. From this, it will follow that $A B C D$ is a circumscribed quadrilateral. (The point of intersection of these bisectors will be equidistant from $A B$ and $A D$, $B C$ and $C D$, as well as from $A B$ and $C D$.)

Take a point $T$ on the extension of segment $E A$ beyond point $A$ such that $A T=A F$, and a point $S$ on the extension of segment $E C$ beyond point $C$ such that $C S=C F$. Since $E T=E A+A F$ and $E S=E C+C F$, it follows from the condition that $E T=E S$.

Consider triangle TFS. The perpendicular bisector of side $T S$ of this triangle is the bisector of angle TES (or angle $B E C$). This follows from the isosceles nature of triangle TES. Similarly, we can prove that the perpendicular bisector of side $T F$ is the bisector of angle $T A F$ (or angle $B A D$), and the perpendicular bisector of side $S F$ is the bisector of angle $C S F$ (or angle $B C D$). Therefore, the specified bisectors intersect at one point - the center of the circumscribed circle of triangle TFS.