Download Inscribed Quadrilateral

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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]