1. ## prove

Let ABC be an acute angled triangle , and let P and Q be two points on side BC.Construct point C1 in such a way that convex quadrilateral APBC1 is cyclic, QC1||CA, and C1 and Q lie on opposite sides of line AB. Construct point B1 in such a way that convex quadrilateral APCB1 is cyclic, QB1 ||BA, and B1 and Q lie on opposite sides of line AC.Prove that points B1, C1,P and Q lie on a circle.

no idea.anyone?

2. Originally Posted by nh149
Let ABC be an acute angled triangle , and let P and Q be two points on side BC.Construct point C1 in such a way that convex quadrilateral APBC1 is cyclic, QC1||CA, and C1 and Q lie on opposite sides of line AB. Construct point B1 in such a way that convex quadrilateral APCB1 is cyclic, QB1 ||BA, and B1 and Q lie on opposite sides of line AC.Prove that points B1, C1,P and Q lie on a circle.
In the attachment, you should be able to see that the three angles marked with a bullet ( $B_1QC,\ ABP,\ AC_1P$) are all equal. Similarly, the three angles marked with a circle ( $C_1QP,\ ACP,\ AB_1P$) are all equal.

The three marked angles at Q (circle, x and bullet) add up to 180º. Also the three angles in the triangle $B_1QC_1$ add up to 180º. But these angles are an x (at Q), a circle plus a star (at $B_1$) and a bullet minus a star (at $C_1$). It follows that the two angles marked with a star ( $PC_1Q$ and $PB_1Q$) must be equal. By the "angles on the same segment" theorem, or rather its converse, it follows that the points $P,\ Q,\ B_1,\ C_1$ lie on a circle.

[It then follows fairly easily from this that the points $B_1,\ A,\ C_1$ are collinear.]