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Intermath | Workshop Support Write-up Title Inscribed Quadrilateral Problem Statement Consider a quadrilateral inscribed in a circle. Construct the angle bisectors for each of the pairs of opposite sides. Prove or disprove: The angle bisectors of the opposite sides of an inscribed quadrilateral meet at right angles. Problem setup Given: Inscribed quadrilateral. Prove or disprove: Angle bisectors are perpendicular. Definitions Inscribed Polygon: A polygon is inscribed in a circle if and only if each of its vertices lie on the circle. Angle Bisector: A segment or ray that shares a common endpoint with an angle and divides the angle into two equal parts. Supplementary Angles: Supplementary angles are two angles whose sum is 180 degrees. Plans to Solve/Investigate the Problem I will create an inscribed quadrilateral with the angle bisectors to verify that select examples are perpendicular. I will then look at the characteristics that would be needed to prove this for every inscribed quadrilateral and attempt to write a proof. Investigation/Exploration of the Problem I drew a circle with Geometer’s Sketchpad (GSP) and constructed four points on the circle. I constructed segments between these points to form a quadrilateral. I then constructed lines between the points to create the intersection of the sides. I constructed the angle bisectors of the two intersections. They met at 90 degrees. I moved the points on the circle and the measurement stayed the same. m ABC = 90.00 A B C m ACB = 90.00 m DEF = 91.01 A m EFG = 126.14 m FGD = 88.99 E m GDE = 53.86 m DEF+m FGD = 180.00 m EFG+m GDE = 180.00 F C D G B Proof Given an inscribed quadrilateral GFED in circle N, we know that the opposite angles, DEF and EGF are supplementary. The external angle of EGF is congruent to DEF . Let this measurement be z. By extending the sides so they intersect at points A and B, two angles, EAF and FBG are formed. Construct the angle bisectors of these two angles and label their intersection, point C. Thus, GBC CBF with a measurement of x and EAC CAF with a measurement of y by definition of angle bisector. Label the intersections of the FBG ’s angle bisector with the quadrilateral as I to the first and H to the second intersection. This creates two triangles, FBI and IBG . Since a triangle has 180 degrees, BIG 180 ( x z ) . Vertical angles are congruent so HIF 180 ( x z ) . By the exterior angle theorem, FIB ( x z ) . Therefore, BFG 180 (2 x z ) since a triangle has 180 degrees. By a definition of linear pair, EFG 2x z . A quadrilateral has 360 degrees. By subtracting the three known angles of quadrilateral EFIH from 360, EHI 180 ( x z ) . Thus, EHI HIF , which makes AIH an isosceles triangle. The angle bisector of the vertex of an isosceles triangle is perpendicular to the base, which is the intersection point C discussed earlier. Extensions of the Problem Consider another polygon inscribed in a circle. Author & Contact Nicole McDowell [email protected]