Question 1162497
<pre>
{{{drawing(400,3200/13,-1,12,-1,7,

triangle(0,0,54/11,0,8,6), triangle(11,0,54/11,0,8,6), 

green(arc(8,6,4,-4,216,298),arc(54/11,0,4,-4,0,63)),

red(arc(8,6,3,-3,245,297),arc(0,0,3,-3,0,37)),
locate(0,0,Q), locate(11,0,R), locate(8,6.5,P), locate(54/11,0,X),
blue(arc(11,0,4,-4,117,180))



)}}}

We are to prove that the angles with the green arcs are equal, 

∠PXR = ∠QPR

given

that the angles with the red arcs are equal,

∠RPX = ∠Q

We use the fact that the three internal angles of a triangle have sum 180°,

For ΔPQR, 

(1)  ∠QPR + ∠R + ∠Q = 180°. Therefore,

(2)  ∠QPR = 180° - ∠R - ∠Q

For ΔRPX, 

(3)  ∠PXR + ∠R + ∠RPX = 180°. Therefore, 

(4)  ∠PXR = 180° - ∠R - ∠RPX 

Since we are given that ∠RPX = ∠Q, the right sides
of (2) and (4) are equal, so their left sides are also equal. 
Therefore,

∠PXR = ∠QPR

Now you can write that up in a two-column proof.

Edwin</pre>