$ABC$ is a triangle in which $L$ is the midpoint of $AB$ and $N$ is a point on $AC$ such that $AN =2CN$. A line thought $L$ parallel to $BN$ meets $AC$ at $M$ prove that $AM=CN$.
Asked
Active
Viewed 18 times
1

Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – user37238 Dec 04 '15 at 10:16

1Also, please explain your tag choice and be consistent in your choice for letters (lowercase/uppercase). – user37238 Dec 04 '15 at 10:17
1 Answers
0
By construction the triangles $ALM$ and $ABN$ are similar and $2AL=AB$, so we have also $2AM=AN$ and, since $AN=2CN$ we have $2AM=2CN \Rightarrow AM=CN$
Emilio Novati
 60,834
 5
 42
 106