Where Do They Concur?

Image link: http://s3.postimg.org/kzkbr62kj/Untitled.png

WLOG we assume the configuration shown in the figure.

Let $X,Y,Z$ be the midpoints of $BC,CA,AB$ respectively, and let $G$ be the centroid of $\triangle ABC.$ Consider the homothety centered at $G$ which maps $A,B,C$ to $X,Y,Z$ respectively (such a homothety exists because $\triangle XYZ \sim \triangle ABC$ ). We shall show that this homothety maps $P$ to $D.$

Note that $\angle PBC = 180^{\circ} - \angle PAC = \angle ACB = \angle AYZ,$ where we have used the facts that $AP \parallel BC$ and $YZ \parallel BC.$ Now, note that $D$ lies on the circle with diameter $AC,$ so $YA = YC = YD.$ From isoceles $\triangle YCD,$ $\angle DYZ= \angle AYZ = \angle PBC.$ Since $AP \parallel BC,$ this homothety maps $P$ to $D.$ Hence, $D,G,P$ are collinear.

Similarly, $E,G,Q$ and $F, G, R$ are also collinear. Therefore, lines $DP,EQ,FR$ are concurrent at the $\fbox{centroid}$ of $\triangle ABC.$

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This problem is inspired by ISL 2011 G4 .

since AD is perpendicular to BC and AP || BC, so AD is perpendicular to AP. Hence PD coincides with the diameter of the circle. Similarly, QE and RF are also along the diameters. Hence they intersect at the centre of the circle i.e. centroid of the triangle ABC.

Observe the diagram. Note that since it is to scale, it just takes some observation and spatial reasoning. We can eliminate some answers: It doesn't look like the circumcenter because it's not in the center of the circumcircle. It can't be the incenter because it doesn't look like one can draw three congruent perpendicular lines to the sides of the triangle. If one visually draws three lines from the vertices to the point, it seems like the opposite edge is bisected. Thus, it must be the $\boxed{\text{centroid}}$