Problem
GEO-B1-M07-P024 Median to the Hypotenuse, Converse
#24
★★★★☆ 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 \(M\) and radius \(MB\).
Since \(M\) is the midpoint of \(BC\), we have \(MB=MC\). By condition \(AM=BM\). Hence \(MA=MB=MC\), and points \(A,B,C\) lie on the circle with centre \(M\). Also, \(BC\) is a diameter of this circle. Angle \(\angle BAC\) stands on the diameter, so \(\angle BAC=90^\circ\).
A strong example: equal distances suggest a circle.