Problem
GEO-B1-M02-P007 A Perpendicular Bisector
#7
★★☆☆☆ Level 2 of 5
Point \(M\) is the midpoint of segment \(BC\). Point \(A\) is chosen so that \(AM\perp BC\). Prove that \(AB=AC\).
Compare right triangles \(ABM\) and \(ACM\).
Since \(M\) is the midpoint of \(BC\), we have \(BM=CM\). Side \(AM\) is common, and angles \(AMB\) and \(AMC\) are right angles. By SAS, triangles \(ABM\) and \(ACM\) are congruent. Therefore \(AB=AC\).
This is the basic perpendicular-bisector property proved through congruent triangles.