## Rhombus in a Cyclic Quadrilateral

In 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?

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny### Discussion

The applet illustrates Problem 86 from R. Honsberger's *Mathematical Morsels*:

What if applet does not run? |

Since ABCD is cyclic, the sum of its opposite angles is 180°, leading to

In triangles AEQ and CGQ angles at Q are equal by the construction, implying the equality of the third pair of angles:

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|

Copyright © 1996-2018 Alexander Bogomolny66597356