Problem
GEO-B2-M01-P008 An Orthic Pair of Angles
In acute triangle \(ABC\), points \(D\) and \(E\) are the feet of the altitudes from \(B\) and \(C\), respectively. Prove that \(\angle ADE\equiv\angle ACB\pmod{180^\circ}\) and \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).
First prove that \(B,C,D,E\) lie on the circle with diameter \(BC\).
Since \(BD\perp AC\), we have \(\angle BDC=90^\circ\). Since \(CE\perp AB\), we have \(\angle BEC=90^\circ\). Therefore \(B,C,D,E\) lie on the circle with diameter \(BC\). Now apply the antiparallel result inside the angle: from cyclicity of \(B,C,D,E\), it follows that \(\angle ADE\equiv\angle ACB\) and \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).
The combination of altitudes, a circle with a diameter, and antiparallel lines is common at the intermediate level.