If $ABCD$ is quadrilateral in which $AB+CD=BC+AD$, prove bisectors of angles meet at point which is equidistant from sides of quadrilateral.

Question

If $ABCD$ is quadrilateral in which $AB+CD=BC+AD$, prove bisectors of angles meet at point which is equidistant from sides of quadrilateral.

My attempt:

I know this is cyclic quadrilateral but don't know how this helps in proving this.

Kindly give me hint.

Answer 1

Hint:

Let lines $AD$ and $BC$ meet at $E$ and inscribe a circle $K$ in a triangle $ABE$ (if $E$ is on halfline $AD$ and $BC$). Then draw a second tangent (different of $BC$) at $C$ to this circle and let is cut line $AD$ at $D'$. Now $ABCD'$ is tangent circle and we have $$AD' + BC = AB+CD'$$

Now what can you conclude for $D'$?

Answer 2

Let $\Phi$ be a circle, which is touched to segments $AB$, $BC$ and $AD$ and $D_1$ be a point on $AD$ such that $CD_1$ is touched to $\Phi.$

Thus, $$AB+CD_1=BC+AD_1,$$ which with $$AB+CD=BC+AD$$ gives $$CD_1-CD=AD_1-AD$$ or $$CD-CD_1=DD_1$$ or $$CD=CD_1+D_1D,$$ which gives $$D\equiv D_1$$ and we are done!