# Geometry problem prove lines are parallel, cyclic quadrilaterals,other parallel lines

• Mar 13th 2010, 10:00 PM
RubiksMan
Geometry problem prove lines are parallel, cyclic quadrilaterals,other parallel lines
Go here for a diagram for this problem: http://i630.photobucket.com/albums/u...tryproblem.jpg

Let ABC be a triangle with orthocentre H and let P be a point on its circumcircle. The line through A parallel to BP meets CH at Q and the line through A parallel to CP meets BH at R.
Prove that QR is parallel to AP

THis question has been bugging me for a very long time. Please put me out of my misery with an easy to follow proof.
• Mar 14th 2010, 02:22 PM
Opalg
In the picture in the attachment, we are given that the blue lines CP and AR are parallel, as are the green lines BP and AQ, and we want to show that the red lines AP and QR are parallel. The proof is by similar triangles, showing that the triangles ABC and ARQ are similar. It then follows that the angle AQR between the green line AQ and the red line QR is equal to the angle ACB. This in turn (angles in the same segment) is equal to the angle APB between the green line BP and the red line AP. Since the green lines are parallel, it follows that the red lines are parallel.

To see that the triangles ABC and ARQ are similar takes a bit of work. Start with the angle AQC, between the green line AQ and the orthogonal CHQ to the line AB. This is the complement of the angle between AB and the green line BP. So $\angle ABP = 90^\circ - \angle AQC$. A similar argument shows that the angle ARH (between the blue line AR and the orthogonal to AC) is related to the angle ACP (between AC and the blue line CP) by $\angle ACP = 90^\circ + \angle ARH$. But the angles ABP and ACP are supplementary (opposite angles of a cyclic quadrilteral), and it follows that $\angle ARH = \angle AQC$.

Now look at the triangle ABR, in which $\angle ABR = 90^\circ - \angle BAC$. The sine rule in this triangle tells you that $\frac{AR}{\sin(90-A)} = \frac{AB}{\sin(\angle ARH)}$ (notice that supplementary angles have the same sine). A similar calculation in the triangle AQC tells you that $\frac{AQ}{\sin(90-A)} = \frac{AC}{\sin(\angle AQC)}$. Since the angles in those equations are equal, it follows that $\frac{AR}{AQ} = \frac{AB}{AC}$.

The final piece of information that we need is that the angles QAR and BAC are equal. (I'll leave out the details of that — it's just a matter of chasing around with angles in the same segment, and angles between parallel lines.) Together with the equality between the ratios of the sides, that completes the proof that the triangles ABC and ARQ are similar.