Problem
GEO-B1-M08-P002 Median in an Isosceles Triangle
#2
★★☆☆☆ Level 2 of 5
In triangle \(ABC\), it is known that \(AB=AC\). Point \(M\) is the midpoint of \(BC\). Prove that \(AM\perp BC\).
Compare triangles \(ABM\) and \(ACM\).
We have \(AB=AC\), \(BM=MC\), and \(AM\) is common. Therefore \(\triangle ABM\cong\triangle ACM\). Hence \(\angle AMB=\angle AMC\). These angles are supplementary, so each is \(90^\circ\), and \(AM\perp BC\).
A classic problem, but in a mixed block the student must choose triangle congruence independently.