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In a triangle $ABC$, $D$ is midpoint of the side $BC$. Through the point $A$, $PQ$ is any straight line. The perpendiculars from the points $B$, $C$ and $D$ on $PQ$ are $BL$, $CM$ and $DN$ respectively. Prove that $DL = DM$

Can someone give me hint as how to use the information, '$D$ is the mid point of BC'? We can use the fact that $BL,DN,CM$ will be parallel to each other

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$\triangle LDN = \triangle MDN (LN=MN, ND - common, \angle LND = \angle MND= 90^{\circ} ) \Rightarrow DL=DM$

$DN - $ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$) enter image description here

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  • $\begingroup$ How is $LN=MN$? $\endgroup$ – Akira Mar 6 '16 at 9:08
  • $\begingroup$ $DN - $ the middle line of the trapezoid $BLMC$ $\endgroup$ – Roman83 Mar 6 '16 at 9:09
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enter image description here $BL||DN||CM$ and $BD=CD$

By intercept theorem $LN=LM$ Clearly $\triangle LND$ and $\triangle MND$ are congruent by S-A-S

So,$ DL=DM $ (Corresponding parts of congruent triangles)

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