point equidistant from all sides of quadrilateral If $ABCD$ is a quadrilateral in which $AB+CD=BC+AD$, Prove that the bisectors of the angles of the quadrilateral meet in a point which is equidistant from the sides of the quadrilateral.
Firstly,I cannot understand why the bisectors would necessarily meet at a point and therefore can"t reach to proof.
 A: I wrote this with a lot of details, but I guess for once one should see a proof of this useful fact. 
Let $ABCD$ be a convex quadrilateral such that $AB + CD = BC +
DA$. Let
$$\angle A = \alpha, \,\, \angle B = \beta, \,\, \angle C =
\gamma, \,\, \angle D = \delta$$ Take the interior angle bisectors
of $ABCD$ through vertices $A$ and $B$, and denote their point of
intersection by $I$. Let $T_{AB}, \, T_{BC}$ and $T_{DA}$ be the
orthogonal projections of point $I$ on the edges $AB, \, BC$ and
$DA$ respectively. Then, since  $AI$ and $BI$ are angle bisectors,
triangles $AIT_{AB}$ and $AIT_{DA}$ are congruent because $\angle
\, IAT_{AB} = \angle \, IAT_{DA} = \alpha/2$, the segment $AI$ is
common for both triangles and $\angle \, AT_{AB}I = \angle \,
AT_{DA}I = 90^{\circ}$. Analogously, triangles $BIT_{AB}$ and
$BIT_{BC}$ are congruent. Consequently
$$IT_{AB} = IT_{BC}=IT_{DA}$$ and $AT_{AB} = AT_{DA}$ as well as
$BT_{AB} = BT_{BC}$. Consequently, if we draw the circle $k_I$
centered at point $I$ and of radius $IT_{AB} = IT_{BC}=IT_{DA}$,
then the points $T_{AB}, \, T_{BC}, \, T_{DA}$ lie on $k_I$.

Choose point $T_{CD}$ on the edge $CD$ so that $CT_{CD} =
CT_{BC}$. Then, if we apply the assumption that $AB + CD = BC +
DA$ then we conclude that $DT_{CD} = DT_{DA}$. Thus the triangles
$DT_{DA}T_{CD}$ and $CT_{BC}T_{CD}$ are isosceles and hence $\angle \, DT_{DA}T_{CD} = 90^{\circ} -
\frac{\delta}{2}$ and $\angle \, CT_{BC}T_{CD} = 90^{\circ}
- \frac{\gamma}{2}$. Therefore, after recalling
that $\angle \, IT_{DA}D = \angle \, IT_{BC}C = 90^{\circ}$, we
deduce that  $\angle \, IT_{DA}T_{CD} =
\frac{\delta}{2}$ and $\angle \, IT_{BC}T_{CD} = \frac{\gamma}{2}$
It follows from analogous arguments that $\angle \, IT_{DA}T_{AB}
= \frac{\alpha}{2}$ and $\angle \, IT_{BC}T_{AB} =
\frac{\beta}{2}$. As a result of all of this angle chasing it
follows that $$\angle \, T_{CD}T_{DA}T_{AB} = \frac{\alpha +
\delta}{2} \,\,\, \text{ and } \,\,\, \angle \, T_{CD}T_{BC}T_{AB}
= \frac{\beta + \gamma}{2}$$ which means that
$$\angle \, T_{CD}T_{DA}T_{AB} +  \angle \, T_{CD}T_{BC}T_{AB} = \frac{\alpha +
\delta}{2} + \frac{\beta + \gamma}{2} = \frac{\alpha + \beta +
\gamma + \delta}{2} = \frac{360^{\circ}}{2} = 180^{\circ}$$ Hence
quad $T_{AB}T_{BC}T_{CD}T_{DA}$ is inscribed in a circle, i.e.
point $T_{CD}$ lies on the circle circumscribed around triangle
$T_{AB}T_{BC}T_{DA}$. However, it was proven that this circle is
$k_I$ with center $I$ so $$IT_{CD} = IT_{AB} = IT_{BC}=IT_{DA}$$
Consequently, triangles $IDT_{DA}$ and $IDT_{CD}$ are congruent
because $IT_{DA} = IT_{CD}$, the segment $ID$ is a common edge and
$DT_{DA} = DT_{CD}$ by construction. This yields that $\angle \, IDT_{DA} =
\angle \, IDT_{CD}$ i.e. $DI$ is the interior angle bisector of vertex $D$
as well as $\angle \, IT_{CD}D = \angle \, IT_{DA}D = 90^{\circ}$,
which means that $CD$ is tangent to circle $k_{I}$ at the point
$T_{CD}$. By analogy we conclude that $CI$ is the interior angle
bisector of vertex $C$.
