## Rhombus in a Cyclic QuadrilateralIn a cyclic quadrilateral ABCD, let P be the intersection of the side lines AB and CD and Q the intersection of AD and BC. What can be said about the bisectors of angles APD and AQB?
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2014 Alexander Bogomolny## DiscussionThe applet illustrates Problem 86 from R. Honsberger's
ABCD is a cyclic quadrilateral with opposite sides extended to meet at P and Q.Prove that the quadrilateral EFGH determined on ABCD by the bisectors of angles P and Q is always a rhombus.
Since ABCD is cyclic, the sum of its opposite angles is 180°, leading to (1)
∠BAD = ∠DCQ.
In triangles AEQ and CGQ angles at Q are equal by the construction, implying the equality of the third pair of angles: (2)
∠AEQ = ∠CGQ.
But ∠CGQ = ∠EGD such that ΔEGP is isosceles, with the base AG. Since, in ΔEGP, FP is the bisector of the angle at P, it is also the perpendicular bisector of the base EG. Similarly, EQ is the perpendicular bisector of FH. This proves that EFGH is a rhombus, as required. ## References-
R. Honsberger,
*Mathematical Morsels*, MAA, 1978, pp. 218-219
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2014 Alexander Bogomolny |

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