Problem
GEO-B1-M02-P028 An Exterior Bisector and a Midpoint
In triangle \(ABC\), point \(D\) lies on the exterior angle bisector of angle \(B\) and inside angle \(A\). It is known that \(\angle BCD=60^\circ\) and \(CD=2AB\). Point \(M\) is the midpoint of segment \(BD\). Prove that \(AM=CM\).
C. Hint 1. Drop the perpendicular from \(D\) to line \(BC\).
D. Hint 2. Use the median to the hypotenuse in right triangle \(BHD\).
E. Full solution.
Let \(H\) be the foot of the perpendicular from \(D\) to \(BC\). Since \(\angle BCD=60^\circ\), right triangle \(DCH\) has a \(60^\circ\) angle at \(C\). Therefore \(CH=\frac{CD}{2}=AB\).
Denote \(\angle CBM=\alpha\). Since \(BD\) is the exterior angle bisector at \(B\), the angle between \(BA\) and \(BM\) is \(180^\circ-\alpha\).
In right triangle \(BHD\), point \(M\) is the midpoint of hypotenuse \(BD\). Hence \(MH=MB\). Therefore \(\angle MHB=\angle HBM=\alpha\), and so \(\angle MHC=180^\circ-\alpha\).
Now compare triangles \(ABM\) and \(CHM\). We have \(AB=CH\), \(BM=HM\), and the included angles are equal: \(\angle ABM=180^\circ-\alpha=\angle CHM\). Thus the triangles are congruent by SAS.
From the congruence, \(AM=CM\).
A. Source analysis. Main objects: an exterior angle bisector, a midpoint, and a right triangle with a \(60^\circ\) angle. The obvious approach is to compare \(AM\) and \(CM\) directly, but the hidden side \(CH=AB\) appears only after dropping a perpendicular. Number of key ideas: 3.
F. Difficulty justification. This is Level 7: a regional-style problem with a non-obvious perpendicular and a final triangle congruence.
G. Check. This is not a one-step exercise: one must construct \(H\), use the median to the hypotenuse, and only then obtain SAS.