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If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.

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In \triangleABC, P, Q and R are the midpoints of the sides BC, CA and AB respectively

Also AD\perp BC

To prove: P, Q, R and D are concyclic

Constructions: Join PR, RD, QD

Proof:

By midpoint theorem

RP || AC

QP || AB

So, ARPQ is a parallelogram

\angleRAQ = \angleRPQ  … (1)          (opposite angles of a ||gm are equal)

Now, in right-angled \triangleADB, R is the midpoint of AB.

The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles because the median equals one-half the hypotenuse.

\because RD = RA

\angleRAD = \angleRDA … (2)   (angles opposite to equal sides in a triangle are equal)

Similarly, in\triangleADQ

\angleDAQ = \angleADQ … (3)

Adding (2) and (3)

\angleRAD + \angleDAQ = \angleRDA +\angleADQ

\angleRAQ = vRDQ

From (1), \angleRAQ =\angleRPQ

Hence, we can see if we consider the arc RQ of a circle then \angleRAQ and\angleRPQ are the angles subtended by it on a circle as these angles are equal.

So, P, Q, R and D are concyclic

Hence proved

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