In Triangle ABC , BM and CN are perpendiculars from points B and C on any line passing through A. If L is the mid-point of BC, prove that ML = NL

Question

I found this question in my textbook and I think this question requires the use of the mid-point theorem. I even tried proving the equality using congruence but couldn't seem to make a headway.

I am in grade 9 so this might seem like a stupid question but please try to exlain in simple terms.

Answer 1

Let $LL' \perp MN$ with $L'$ on $MN$. Then $LL'$ is the midline in the trapezoid $BMNC$, thus $L'$ is the midpoint of $MN$.

This shows that $ML'=L'N$.

By Pytagorean theorem

$$ML^2=ML'^2+L'L^2=L'N^2+L'L^2=NL^2$$