100 POINTS!!!! I do not understand what so do here!!!!

Answer:
[tex]\angle LMK\sim \triangle KMJ \sim \triangle LKJ[/tex]
Step-by-step explanation:
In the right triangle shown, two right triangles are inscribed in a third right triangle, with all three triangles sharing an altitude. Thus, all three right triangles are similar.
The second portion asks to draw them so that their corresponding sides and angles have the same orientation. I can't demonstrate that here but here are some tips:
Here is an example of the drawing portion from the CK-12 Foundation:
Answer:
Solution given:
In triangle LMK and triangle LKJ
<LMK=<JMK[ being right angle]
<MLK=<JLK[common angle]
<MKL=<LJK[remaining angle]
so
∆MLK~ ∆MLJ[By A.A.A.axiom]
∆MLK is similar to ∆MLJ