Problem
GEO-B1-M02-P022 Points on the Sides and a Median
#22
★★★★☆ Level 4 of 5
In triangle \(ABC\), median \(AM\) is drawn. Points \(P\) and \(Q\) are chosen on sides \(AB\) and \(AC\) so that \(AP=AQ\) and \(BP=CQ\). Prove that \(PM=QM\).
First prove that \(AB=AC\). Then median \(AM\) becomes an angle bisector.
Since \(P\) lies on \(AB\) and \(Q\) on \(AC\), we have \(AB=AP+PB\) and \(AC=AQ+QC\). From \(AP=AQ\) and \(BP=CQ\), it follows that \(AB=AC\). Hence triangle \(ABC\) is isosceles, and median \(AM\), drawn to base \(BC\), is the angle bisector of angle \(A\). Now consider triangles \(APM\) and \(AQM\). They have \(AP=AQ\), common side \(AM\), and \(\angle PAM=\angle MAQ\). By SAS, the triangles are congruent, so \(PM=QM\).
A good closing problem: first derive isoscelesness, then apply triangle congruence.