In both cases, you have to work on the angle between the sides you found out to be equal.
Diagram 1:
You can show that the angle PCB is the same as ACR. To prove this, you can think of angles:
[tex]P\hat{C}B=P\hat{C}A + A\hat{C}B[/tex]
and
[tex]A\hat{C}R=B\hat{C}R+A\hat{C}B[/tex]
So, angle [tex]A\hat{C}B[/tex] is common, and both [tex]P\hat{C}A[/tex] and [tex]B\hat{C}R[/tex] are right angles.
So, angles PCB and ACR are both 90° plus a common term (ACB), and thus they're equal.
Diagram 2:
We will work very similarly here: in this case, you have
[tex]A\hat{B}G = A\hat{B}C-G\hat{B}C[/tex]
and
[tex]C\hat{B}E=G\hat{B}E-G\hat{B}C[/tex]
Again, both [tex]A\hat{B}C[/tex] and [tex]G\hat{B}E[/tex] are right angles, so we have
[tex]A\hat{B}G = 90-G\hat{B}C[/tex]
and
[tex]C\hat{B}E=90-G\hat{B}C[/tex]
So, both [tex]A\hat{B}G[/tex] and [tex]C\hat{B}E[/tex] are 90 minus a common term, so they are equal.
