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Given $ABCD$ - Trapezoid $M$ is mid point of $AD$, $N$is mid point of $BC$. The angle bisectors of $\angle BAD$ and $\angle ADC$ are intersecting at point $P$ prove that this point lies on $MN$. I have no idea from where to start. Here is diagram(sort of)

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Since $\widehat{A}$ and $\widehat{D}$ are supplementary angles, $DPA$ is a right triangle and $PN=NA=ND$.

Let $E$ be $DP\cap AB$. By angle chasing, $EA=AD$. $DNP$ and $DAE$ are similar triangles, hence $NP\parallel AE$ and $NP=\frac{1}{2} AE$ by Thales' theorem.

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