Answer :

** 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.

__Proof:__

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° …[1]

⇒ ∠POQ + ∠C = 180° …[2]

⇒ ∠POR + ∠B = 180° …[3]

Also, By angle sum property of triangle

∠A + ∠B + ∠C = 180° …[4]

Now, Adding [1] [2] and [3]

⇒ ∠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.

Rate this question :

In the picture, tKerala Board Mathematics Part-1

The two circles bKerala Board Mathematics Part-1

In the picture beKerala Board Mathematics Part-1

In the picture, pKerala Board Mathematics Part-1

In the picture, bKerala Board Mathematics Part-1

Calculate the angKerala Board Mathematics Part-1

Prove that any exKerala Board Mathematics Part-1

Prove that a paraKerala Board Mathematics Part-1

Prove that a non-Kerala Board Mathematics Part-1