Problem
GEO-B1-M05-P021 A Hidden Circle on the Sides of a Triangle
#21
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in AB\) and \(E\in AC\). It is known that \(\angle ADE=\angle ACB\). Prove that points \(B,C,D,E\) lie on one circle.
Since \(D\) lies on \(AB\), angle \(\angle ADE\) can be related to angle \(BDE\).
Points \(A,D,B\) are collinear, so \(\angle BDE=180^\circ-\angle ADE\). By condition \(\angle ADE=\angle ACB\). Since \(E\) lies on \(AC\), angle \(\angle BCE=\angle BCA=\angle ACB\). Thus \(\angle BDE+\angle BCE=180^\circ\). The sum of opposite angles of quadrilateral \(BCDE\) is \(180^\circ\), therefore \(B,C,D,E\) lie on one circle.
A strong problem on the converse cyclicity criterion: the angle equality is hidden through a straight angle.