The Figure on the right shows two circles with centers at points R and S, 

tangent externally at point P and touching the given straight line at points A and B.   


We need to prove the statement that the triangle BPA  is right angled triangle.

  
Let  < 1  be the angle PAB;  < 2  be the angle PBA;  < 3  be the angle BPA;

     < 4  be the angle PAR;  < 5  be the angle APR;  < 6  be the angle PBS;

     < 7 be the angle BPS.


Notice that  < 4 = < 5, since the triangle ARP is isosceles.

Similarly,   < 6 = < 7, since the triangle BSP is isosceles.

 
  



 
 
Also notice that RPS is a straight line, because the segments RP and SP are perpendicular radii to the common tangent line 
to the two circles at their touching point P.

We then have

        < 1 + < 4 = 90°,
        < 2 + < 6 = 90°,

which implies  < 1 + < 2 = 180° - (< 4 + < 6) = 180° - (< 5 + < 7) = < 3.   (1)


Thus we have these two equalities

     < 1 + < 2 - < 3 = 0      (2)     ( same as (1) ),   and

     < 1 + < 2 + < 3 = 180°   (3)     ( as the sum of interior angles of the triangle BPA).

By adding equations (2) and (3), you get  < 1 + < 2 = 90°,  which immediately implies that < 3 = 90°.