Problem 1
We're given that QE = RW and QW = RE. This is given both in terms of symbols, and also visually in terms of the tickmarks. We have two pairs of congruent sides.
The third pair of sides is the overlapping shared segment of segment WE. So WE = WE by the reflexive property.
From here we use the SSS congruence property to conclude the proof that triangle QWE = triangle REW
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Problem 2
Segments BC and AE bisect each other. The term "bisect" means "to cut in half".
BC bisecting AE means AD = DE
AE bisecting BC means BD = DC
So we have two pairs of congruent sides. So far, we're on a similar path compared to problem 1. This time however, we'll use a pair of congruent angles instead of the third pair of sides.
Specifically we'll use the fact that
angle ADB = angle CDE
by the vertical angle theorem
Once we know that, we then use the SAS congruence theorem to prove triangle ABD = triangle ECD
To be honest, I'm not sure what you meant by "median", but we don't use any triangle medians here (none of the triangles shown have any medians drawn). You might be thinking of a previous problem.
