In triangle ABC, D is the mid-point of BC. If DL perpendicular to AB and DM perpendicular to AC such that DL = DM, then the triangle will be:

$D$ is the mid-point of $BC$.

$DL\perp AB$, and $DM\perp AC$.

$DL=DM$.   

Concept Used:  

If two triangle have one side , hypotenuse and right angle equal to each other,

then both triangles are congruent through RHS congruency. 

In $$\triangle BDL$$ and $$\triangle CDM$$ 

DL = DM 

$$\angle L = \angle M = 90^\circ$$ 

BD = CD ( D is mid point of BC) 

So, $$△BDL \cong△CDM$$ ( RHS congruency)

thus BL = CM .... (1)

In $$\triangle ALD$$ and $$\triangle AMD$$ 

AD = AD (common)

​$$\angle L = \angle M = 90^\circ$$ 

DL = DM  

therefore, $$△ALD≅△AM D$$ (RHS congruency)

thus, AL = AM .... (2)

from (1) and (2) 

we get  AB = AC 

Therefore, $$\triangle ABC$$ is an isosceles triangle. ​