# For a given triangle, prove that $DL=DM$

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

$\triangle LDN = \triangle MDN (LN=MN, ND - common, \angle LND = \angle MND= 90^{\circ} ) \Rightarrow DL=DM$
$DN -$ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$)
• How is $LN=MN$? – Akira Mar 6 '16 at 9:08
• $DN -$ the middle line of the trapezoid $BLMC$ – Roman83 Mar 6 '16 at 9:09
$$BL||DN||CM$$ 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)