Please forgive my weak English.
Let's say $triangle ABC$ is the original triangle, $D$ is the intersection of altitude through $B$ and median through $A$,
$E$ is the intersection of altitude through $B$ and angle bisector through $C$,
$F$ is the intersection of median through $A$ and angle bisector through $C$,
1. Draw straight lines $DE$ (altitude), $DF$ (median) and $EF$ (angle bisector),
2. Draw orthogonal line of $EF$ through $E$(let's say it is $l$, hence angle between $l$ and straight line $DE$ is $\frac{C}{2}$)
3. Take point $C$ on straight line $EF$ externally so that $EC>FC$, take two point $X$, $Y$ at different side of $EC$ so that angle $ECX=ECY=$angle between $l$ and straight line $DE$($=\frac{C}{2}$)
4. Straight line $CX$ intersects $DF$ at $A$ and straight line $CY$ intersects $DE$ at $B$, triangle $ABC$ is the triangle that we are finding.