Question

In ΔАВС, M and N are the mid-points of AB and AC respectively and R be any point on BC. Use intercept theorem, prove that MN bisects AR.

Answer 1

Given:

In triangle ABC, M and N are mid-points of AB and AC respectively.

R is any point on BC.

MN is therefore the segment joining the mid-points, so MN || BC.

To Prove:

MN bisects AR i.e., MN meets AR at its midpoint.

Proof [Step-wise]:

1. Let MN meet AR at point P.

2. Since M and N are mid-points of AB and AC, by the midpoint or mid-segment theorem MN || BC.

Because R lies on BC, MN is parallel to BR and CR.

3. Consider triangle ABR.

The line MN meets AB at M and AR at P and is parallel to BR.

4. By the Intercept Basic Proportionality Theorem applied to triangle ABR with the line through M parallel to BR, we have `(AM)/(MB) = (AP)/(PR)`.

5. M is the midpoint of AB.

So, AM = MB.

Hence, `(AM)/(MB) = 1`.

6. From steps 4 and 5,

`(AP)/(PR) = 1`

So, AP = PR.

7. Therefore, P is the midpoint of AR; equivalently MN bisects AR.

MN bisects AR the point of intersection of MN and AR divides AR into two equal parts.