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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.

MayankJain
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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

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

N. S.
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  • That's nice! +1 for that, too. – Michael Hoppe Nov 11 '13 at 18:11
  • @N.S. Sir if you dont mind can u please explain how $LL′$ is the midline in the trapezoid $BMNC$ and $L′$ is the midpoint of $MN$. – Snehil Sinha Jan 19 '15 at 15:45
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    @SNEHILSINHA For simplicity let $C'$ be the intersection of $LL'$ with $MC$ and $B'$ the intersection of $LL'$ with $BN$. Then $LC'$ is parallel to $BM$ as both are are perpendicular on $MN$. Since $L$ is the midpoint of $BC$ it follows that $LC'$ is midline in the triangle $BCM$. Therefore $C'$ is the midle of $MC$ and $B'C'$ is parallel to the bases.... – N. S. Jan 20 '15 at 00:18
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    @SNEHILSINHA What I prove in this comment is the well known fact that in a trapezoid, the parallel to the bases through the midpoint of a diagonal is the midline. – N. S. Jan 20 '15 at 00:21
  • @N.S. thanx sir – Snehil Sinha Jan 20 '15 at 06:56