For the first one, you did good. I will just suggest a couple things.
Statement Reason
JK ≅ LM Given
<JKM ≅ < LMK Given (You did both of these steps so well done.)
MK ≅ MK Reflexive Property (Your angle pair is congruent but isn't one of the interior angle of the triangles you are trying to prove.)
ΔJMK ≅ ΔLKM SAS
Problem 2: (You also have a lot of great stuff here.)
Statement Reason
DE ║ FG Given
DE ≅ FG Given
<DEF≅<FGH Given
<EDF≅<GFH Corresponding Angles (You don't need to know that F is the midpoint but you got corresponding angle pair which is correct.)
ΔEDF≅ΔGFH ASA
<DFE≅<FHG CPCTC
You didn't provide the "given point", but I assume you're capable of plugging it in.
Answer:
The answer is A I just looked it up on mathaway and it is never wrong
Step-by-step explanation:
Standard form always has the largest coefficients first
Answer:
Yes by SAS
Step-by-step explanation:
If you look at the triangles they have 2 side lenght in common (and since it is a right triange they have all the sides in commmon) and they share an angel
In other words 2 sides are conurent with an angle betwene them.
You would need to rotait the shape 90 degrese.
You keep adding 2 and 1/3 to get your answer
