Given: In the picture, points P, Q, R are marked on the sides BC, CA, AB of Δ ABC and the circumcircles, of Δ AQR, Δ BRP, Δ CPQ are drawn.
To Prove: All the circles pass through a common point.
Let us assume a point O, which passes through all the circles.
and Join OP, OQ and OR.
If we prove that such a point O exists, then we are done.
Now, OQAR, OPCQ and OPBR are cyclic quadrilaterals, and in a cyclic quadrilateral, sum of any pair of opposite angles is 180°.
⇒ ∠QOR + ∠A = 180° …
⇒ ∠POQ + ∠C = 180° …
⇒ ∠POR + ∠B = 180° …
Also, By angle sum property of triangle
∠A + ∠B + ∠C = 180° …
Now, Adding   and 
⇒ ∠QOR + ∠POQ + ∠POR + ∠A + ∠C + ∠B = 540°
⇒ ∠POQ + ∠QOR + ∠POR + 180° = 540° [From 4]
⇒ ∠POQ + ∠QOR + ∠POR = 360°
Now, sum of all the angles around O is 360°,
⇒ O exists and is common to all circles.
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