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  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by nh149 View Post
    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 180. Also the three angles in the triangle B_1QC_1 add up to 180. 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.]
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