Let the perpendicular from point A on side BC cuts PQ at O.
In triangle `DeltaABC` , P and Q are midpoints of sides AB and AC respectively. According to triangle midsegment theorem, PQ||BC. OM is the transversal.
Hence corresponding angles <AOP and <BMA are congruent and each are right angles. Therefore, <POM is also a right angle.
Consider triangles `DeltaAPO` and `DeltaABM` ,
<AOP `cong ` <AMB (both right angles)
<BAM is common to both.
Remaining angles must be equal. So, they are similar.
So, O is the midpoint of AM, i.e. AO=OM.
Now consider right triangles `DeltaAPO` and `DeltaPOM`
< AOP `cong` <POM (both right angles)
OP is the common side.
Triangles `DeltaAPO cong DeltaPOM` (SAS congruence)
Angles opposite to equal sides must be congruent.
So, <APO `cong ` <OPM
Finally, for the quadrilateral PQRM,
=(`90^o` -<OPM)+`90^o` +<PQR
=`180^o` -<PQR+<PQR (AB||QR, PQ is the transversal, and <APO and <PQR are alternate interior angles, hence congruent)
For the quadrilateral PQRM, opposite angles (<PMR and <PQR) are supplementary. Hence, PQRM is a cyclic quadrilateral.