Question 1115257
<pre><TABLE>
  <TR>
  <TD>

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.

 </TD>
  <TD>
{{{drawing( 480, 300,  -0.5, 15.5, -1.0, 9.0,

            line( 0, 0, 15, 0),

            circle (3,   3,   0.1),
            locate( 2.5, 3.2, R),

            circle (3,   3,  3),


            circle (10,   4,   0.1),
            locate( 10.2, 4.2, S),

            circle (10,  4,  4),

            line(3,  3,  3, 0),

            line(10, 4, 10, 0),

            line(3, 3,  10, 4),

            locate(2.8, -0.2, A),

            locate(9.8, -0.2, B),

        red(line(3,   0,  6, 3.3)),

        red(line(10,  0,  6, 3.3)),

            locate(6.2,   4.1, P)
)}}}

 </TD>
 </TR>
</TABLE>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°.  
</pre>

QED.  &nbsp;&nbsp;Proved and solved.