MathDB

1

Part of 2006 QEDMO 3rd

Problems(1)

|EAC| * |EBD| = |EAB| * |ECD| + |EBC| * |EDA|

Source: 3rd QEDMO 2006, problem 1, originally Praxis der Mathematik problem P151

4/14/2006
Peter is a pentacrat and spends his time drawing pentagrams. With the abbreviation ∣XYZ∣\left|XYZ\right| for the area of an arbitrary triangle XYZXYZ, he notes that any convex pentagon ABCDEABCDE satisfies the equality ∣EAC∣⋅∣EBD∣=∣EAB∣⋅∣ECD∣+∣EBC∣⋅∣EDA∣\left|EAC\right|\cdot\left|EBD\right|=\left|EAB\right|\cdot\left|ECD\right|+\left|EBC\right|\cdot\left|EDA\right|. Guess what you are supposed to do and do it.
geometrycircumcirclegeometry proposed