Problem
GEO-B1-M05-P030 Two Hidden Cyclicities
#30
★★★★★ Level 5 of 5
In acute triangle \(ABC\), altitudes \(BD\) and \(CE\) meet at point \(H\). Prove that \(\angle ADE=\angle AHE\).
First prove that \(A,D,H,E\) lie on one circle.
Since \(BD\perp AC\), we have \(\angle ADH=90^\circ\). Since \(CE\perp AB\), we have \(\angle AEH=90^\circ\). Thus \(A,D,H,E\) lie on the circle with diameter \(AH\). In this circle, angles \(\angle ADE\) and \(\angle AHE\) stand on the same chord \(AE\). Therefore \(\angle ADE=\angle AHE\).
Final problem of the module: first build the circle, then use the main pattern of equal angles on one chord.