angles in triangles

1.
PQ = PR (given)
PQR = PRQ (base s of isos. D)
Let PQR = PRQ = 2y

QX is the bisector of PQR. (given)
Hence, PQX = XQR = y

QXR + XQR + PRQ = 180o ( sum of D)
105o + y + 2y = 180o
3y = 75o
y = 25o

PQX + QPR = 105o (ext. of D)
y + QPR = 105o
25o + QPR = 105o
QPR = 80o


2.
ABD = AED = 30o (given)
AB = AE (sides to equal s)
BD = DE
DABD DAED (SAS)
BAC = EAC (corr. s of congr. Ds)

In DABE:
BAC + EAC + ABD + AED = 180o ( sum of D)
BAC + BAC + 30o + 30o = 180o (substitution)
2BAC = 120o
BAC = 60o

In DABC:
BAC + ABC + ACB = 180o ( sum D)
60o + (30o + 30o) + ACB = 180o (substitution)
ACB = 60o

Since BAC = ABC = ACB = 60o
DABC is an equilateral triangle.
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