I am not sure about my solution. Please rectify me if it is wrong. Triangle $PQM$ is the medial triangle of triangle $ABC$. so all 4 triangles are congruent. In triangle $BPM$ $<BPM=90$. so by simplicity $<PAQ=90$

geosolver26 wrote: I am not sure about my solution. Please rectify me if it is wrong. Triangle $PQM$ is the medial triangle of triangle $ABC$. so all 4 triangles are congruent. In triangle $BPM$ $<BPM=90$. so by simplicity $<PAQ=90$ Although this is certainly the case when $\angle A=90^{\circ}$, when we're not in this special case, $\triangle MPQ$ is not the medial triangle of $ABC$ (draw a diagram for a small $\angle A$ to see this clearer), because $P$ and $Q$ would be feet of the perpendiculars from $M$ to $AB$, $AC$ and those are midpoints only if $\angle A=\angle 90^{\circ}$. A lot more is required, though your post gives you valuable insight.