Bariz - Problem

Let $N$ be the midpoint of $DE$ and $I$ be the incenter of $ABC$. Then $NI\perp BC$. See here for the proof. 1. Construct the midpoint $N$ of $DE$ and draw the line $TN$. 2. Construct the corresponding arc $ED$ with measure of $90^\circ$ and mark the point $I$ of its intersection with $NT$. 3. Draw the line $a$ through $T$ perpendicular to $IN$ and mark the points $B, C$ of its intersection with $DI, EI$ respectively. 4. Mark the point $A$ of the intersection of $BE$ and $CD$. Hence the triangle is constructed.