Question

In Triangle ABC. P, Q and R are the midpoints of sides AB, AC and BC. side AS = side BC. Prove that: Quadrilateral PQRS is cyclic

Answer 1

Answer:

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Step-by-step explanation:

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 <BSA are congruent and each are right angles. Therefore, <POM is also a right angle.

Consider triangles DeltaAPO and DeltaABS ,

<AOP cong  <ASB (both right angles)

<BAS is common to both.

Remaining angles must be equal. So, they are similar.

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

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

Now consider right triangles DeltaAPO and DeltaPOS

< AOP cong <POS (both right angles)

AO=OS

OP is the common side.

Triangles DeltaAPO cong DeltaPOS (SAS congruence)

Angles opposite to equal sides must be congruent.

So, <APO cong  <OPS

Finally, for the quadrilateral PQRS,

<PSR+<PQR=<PSO+<OSR+<PQR

=(90^o -<OPS)+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 PQRS, opposite angles (<PSR and <PQR) are supplementary.

Hence, PQRS is a cyclic quadrilATERAL.

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