Problem
GEO-B1-M02-P006 Three Properties of One Line
#6
★★☆☆☆ Level 2 of 5
In isosceles triangle \(ABC\), \(AB=AC\). Point \(M\) is the midpoint of base \(BC\). Prove that \(AM\) is the angle bisector of angle \(A\) and an altitude of the triangle.
First prove that triangles \(ABM\) and \(ACM\) are congruent.
By SSS, triangles \(ABM\) and \(ACM\) are congruent: \(AB=AC\), \(BM=CM\), and \(AM\) is common. Therefore \(\angle BAM=\angle MAC\), so \(AM\) is an angle bisector. Also \(\angle AMB=\angle AMC\). These angles are adjacent, so both are \(90^\circ\). Hence \(AM\perp BC\), so \(AM\) is an altitude.
A key basic theorem of the module; ask for a full proof via SSS.