See Explanation
Step-by-step explanation:
1. By interior angle sum Postulate of a triangle.
[tex] m\angle a+ m\angle y +m\angle z= 180\degree.... (1)\\\\
m\angle a+ m\angle w +m\angle x= 180\degree.... (2)\\\\[/tex]
From equations (1) & (2), we find:
[tex] m\angle a+ m\angle y +m\angle z= m\angle a+ m\angle w +m\angle x\\\\
m\angle y +m\angle z= m\angle w +m\angle x\\[/tex]
Hence proved
2. In [tex] \triangle PQS, \: \angle QSR [/tex] is exterior angle.
Therefore, by remote interior angle theorem, we have:
[tex] m\angle QPS + m\angle PQS= m\angle QSR\\\\
x + m\angle PQS= 2x\\\\
m\angle PQS= 2x-x\\\\
m\angle PQS = x.... (1)\\\\
m\angle QPS = x.... (given).... (2)\\[/tex]
From equations (1) & (2), we find:
[tex] m\angle PQS = m\angle QPS\\\\
\therefore \angle PQS \cong \angle QPS\\\\
\therefore PS \cong QS\\(sides\: opposite \: to\: congruent \:\angle s) \\\\[/tex]
[tex] \therefore \triangle PQS [/tex] is an isosceles triangle.
Thus proved.
