Problem
GEO-B1-M05-P015 Two Altitudes
#15
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(D\) is the foot of the altitude from \(B\) to \(AC\), and point \(E\) is the foot of the altitude from \(C\) to \(AB\). Prove that \(B,C,D,E\) lie on one circle.
Prove that \(\angle BDC=\angle BEC=90^\circ\).
Since \(BD\perp AC\) and \(D\in AC\), we get \(\angle BDC=90^\circ\). Since \(CE\perp AB\) and \(E\in AB\), we get \(\angle BEC=90^\circ\). Thus points \(D\) and \(E\) see segment \(BC\) under a right angle. Therefore they lie on the circle with diameter \(BC\), so \(B,C,D,E\) are cyclic.
A classic building block for harder altitude problems.