Problem
GEO-B1-M02-P013 An Extended Median
#13
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). On ray \(AM\) beyond \(M\), point \(P\) is chosen so that \(MP=AM\). Prove that \(BP=AC\) and \(CP=AB\).
Use vertical angles at the intersection of lines \(AP\) and \(BC\).
We have \(AM=MP\) and \(BM=CM\). Angles \(AMB\) and \(PMC\) are vertical, so they are equal. By SAS, \(\triangle AMB=\triangle PMC\), hence \(AB=PC\). Similarly, angles \(AMC\) and \(PMB\) are vertical, with \(AM=MP\) and \(CM=BM\). Thus \(\triangle AMC=\triangle PMB\), so \(AC=PB\).
A classic way to “double a median” using congruent triangles.