Problem
GEO-B1-M08-P023 A Circle From Equal Distances
#23
★★★★☆ Level 4 of 5
In triangle \(ABC\), point \(M\) is the midpoint of \(BC\), and \(AM=BM\). Prove that \(\angle BAC=90^\circ\).
Draw the circle with centre \(M\).
Since \(M\) is the midpoint of \(BC\), \(BM=CM\). By condition \(AM=BM\), hence \(AM=BM=CM\). Therefore points \(A,B,C\) lie on the circle with centre \(M\), and \(BC\) is a diameter. Angle \(\angle BAC\) stands on the diameter, so it is right.
The problem looks like midpoints, but is solved by a circle.