# In the Fig. ,ABC is a triangle and P, Q and R are the mid points. AM is perpendicular to BC. Prove that PQRM is a cyclic Quadrilateral.

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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.

Therefore, AP/AB=AO/AM=1/2

So, O is the midpoint of AM, i.e. AO=OM.

Now consider right triangles `DeltaAPO` and `DeltaPOM`

< AOP `cong` <POM (both right angles)

AO=OM

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,

<PMR+<PQR=<PMO+<OMR+<PQR

=(`90^o` -<OPM)+`90^o` +<PQR

=`180^o` -<APO+<PQR

=`180^o` -<PQR+<PQR (AB||QR, PQ is the transversal, and <APO and <PQR are alternate interior angles, hence congruent)

`=180^o`

For the quadrilateral PQRM, opposite angles (<PMR and <PQR) are supplementary. Hence, PQRM is a cyclic quadrilateral.