ad is the median of triangle abc. m,n are points on ab,ac. ad intersection mn =k. if mk=kn then prove that mn is parallel to bc.

To prove that line segment MN is parallel to BC, we can make use of similar triangles. Here's how we can approach the proof:

1. Let's start by drawing a diagram of triangle ABC with AD as the median, and points M and N on sides AB and AC respectively. The intersection of line AD and line MN is point K.

2. Since AD is the median, it will divide side BC into two equal segments. Let's label the point of intersection of AD and BC as P, such that BP = PC.

3. Since MK = KN (given in the problem), this means that point K is equidistant from points M and N.

4. Now let's consider triangles APK and CPN. By the midpoint property of a triangle, we know that AP = PC and KP = KP (common). Therefore, triangles APK and CPN are congruent by the side-side-side (SSS) congruence criterion.

5. Congruent triangles have corresponding angles equal. Therefore, angle PAK = angle PCN.

6. Now let's consider triangles APK and KDN. Since AD is the median, it will divide side BC into two equal segments. Therefore, BD = DC.

7. By the midpoint property, we know that BP = PD and KP = KD. Therefore, triangles APK and KDN are congruent by the side-side-side (SSS) congruence criterion.

8. Congruent triangles have corresponding angles equal. Therefore, angle APK = angle KDN.

9. Now let's consider the alternate interior angles formed by lines MN and BC. Angle PCN and angle KDN are corresponding angles, and we have already proved that angle PAK = angle PCN and angle APK = angle KDN.

10. Since alternate interior angles are congruent when lines are parallel, we can conclude that MN is parallel to BC.

Therefore, using the concept of congruent triangles and alternate interior angles, we have proved that MN is parallel to BC.