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Solutions - University of Regina
... Also the quadrilateral ABP C is cyclic (because P has been chosen on the circumcircle of ABC). We deduce that ∠BP C + ∠BAC = π ⇒ ∠BP C = π − ∠BAC. So ∠U P W = ∠BP C, as desired. In conclusion, we have proved that ∠BV U = ∠CV W . This implies that the points U, V , and W are collinear. 2*. As the hin ...
... Also the quadrilateral ABP C is cyclic (because P has been chosen on the circumcircle of ABC). We deduce that ∠BP C + ∠BAC = π ⇒ ∠BP C = π − ∠BAC. So ∠U P W = ∠BP C, as desired. In conclusion, we have proved that ∠BV U = ∠CV W . This implies that the points U, V , and W are collinear. 2*. As the hin ...