Track / Stage 5 / 376 of 400 #976 of 2000

Problem 976

AIME late

Geometry Difficulty 6.0 Prove it

## Task 3 - 070723

Given is a triangle $\triangle A B C$ . $M$ is the midpoint of side $A C$ . The line parallel to side $A B$ through point $M$ intersects side $B C$ at point $N$ .

Prove that $N$ is the midpoint of side $B C$ !

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

The parallel line $p$ to $BC$ through $M$ intersects $AB$ at $P$ . That this parallel line $p$ actually intersects the segment $AB$ follows as such:

By $p$ , the plane in which $\triangle ABC$ lies is divided into two half-planes, one of which contains the line through $B$ and $C$ , and the other contains $A$ , because otherwise $AC$ would not be intersected by $p$ . Therefore, $p$ also intersects the segment $AB$ .

Then the following holds:

(1) $\angle ABC = \angle MNC = \angle APM$ , $\angle CAB = \angle CMN$ (as corresponding angles at intersected parallels)

(2) $\triangle APM = \triangle MNC$ (by the congruence theorem sas). From this, it follows that $MP = CN$ .

(3) Quadrilateral $BNMP$ is a parallelogram (by construction). From (2) and (3), the claim follows.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.