Problem
GEO-B2-M01-P005 An Antiparallel Chord
#5
★★★☆☆ Level 3 of 5
In triangle \(ABC\), points \(D\) and \(E\) lie on sides \(AB\) and \(AC\), respectively. It is known that \(B,C,D,E\) lie on one circle. Prove that \(\angle ADE\equiv\angle ACB\pmod{180^\circ}\) and \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).
In circle \(B,C,D,E\), compare angles standing on chords \(BE\) and \(CD\).
Since \(B,C,D,E\) are cyclic, \(\angle BDE\equiv\angle BCE\pmod{180^\circ}\). But \(BD\) lies on line \(BA\), and \(CE\) lies on line \(CA\), so \(\angle ADE\equiv\angle ACB\pmod{180^\circ}\). Similarly, from \(\angle BED\equiv\angle BCD\), we get \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).
This is an important pattern: a cyclic quadruple inside an angle gives an antiparallel.