In convex quadrilateral ABCD, AB = CD, <ABC = 77 degrees and <BCD = 150 degrees.
Let P be the intersection of the perpendicular bisectors of line segment BC and line segment AD.
Find <BPC.
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To find angle BPC in convex quadrilateral ABCD, we can make use of the property that the intersection of the perpendicular bisectors of two sides of a quadrilateral lies on the circumference of the circumcircle of the quadrilateral.
Here's how we can find angle BPC step by step:
Step 1: Draw a diagram of the given information. Label points A, B, C, and D as specified – AB = CD, ∠ABC = 77°, and ∠BCD = 150°. Also, draw the perpendicular bisectors of segments BC and AD and label their intersection point as P.
Step 2: Identify the circumcircle of the quadrilateral ABCD. The circumcircle is a unique circle passing through all four vertices of the quadrilateral.
Step 3: Observe that the perpendicular bisector of a chord in a circle passes through the center of the circle. Since point P is the intersection of the perpendicular bisectors of segments BC and AD, it lies on the center of the circumcircle.
Step 4: Draw the line segment BP. Since P lies on the center of the circumcircle, triangle BPC is an isosceles triangle with BP = PC.
Step 5: Using the property of opposite angles in a cyclic quadrilateral, ∠ABC + ∠BCD = 180°. In our case, we have ∠ABC = 77° and ∠BCD = 150°. Therefore, ∠ABD = 180° - (77° + 150°) = -47°.
Step 6: Since ∠ABD is an exterior angle of triangle BPC, ∠BPC = ∠ABD + ∠ABC = -47° + 77° = 30°.
Therefore, <BPC = 30 degrees.