document.write( "Question 1115257: If two circles, tangent externally at P, touch a given line at points A and B, prove that angle BPA is a right angle. Thank you. \r
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Algebra.Com's Answer #730179 by ikleyn(52788) $\"\"$ $\"About$

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document.write( "The Figure on the right shows two circles with centers at points R and S, \r\n" );
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document.write( "tangent externally at point P and touching the given straight line at points A and B.   \r\n" );
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document.write( "We need to prove the statement that the triangle BPA  is right angled triangle.\r\n" );
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document.write( "Let  < 1  be the angle PAB;  < 2  be the angle PBA;  < 3  be the angle BPA;\r\n" );
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document.write( "     < 4  be the angle PAR;  < 5  be the angle APR;  < 6  be the angle PBS;\r\n" );
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document.write( "     < 7 be the angle BPS.\r\n" );
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document.write( "Notice that  < 4 = < 5, since the triangle ARP is isosceles.\r\n" );
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document.write( "Similarly,   < 6 = < 7, since the triangle BSP is isosceles.\r\n" );
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document.write( " Also notice that RPS is a straight line, because the segments RP and SP are perpendicular radii to the common tangent line \r\n" );
document.write( "to the two circles at their touching point P.\r\n" );
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document.write( "We then have\r\n" );
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document.write( "        < 1 + < 4 = 90°,\r\n" );
document.write( "        < 2 + < 6 = 90°,\r\n" );
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document.write( "which implies  < 1 + < 2 = 180° - (< 4 + < 6) = 180° - (< 5 + < 7) = < 3.   (1)\r\n" );
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document.write( "Thus we have these two equalities\r\n" );
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document.write( "     < 1 + < 2 - < 3 = 0      (2)     ( same as (1) ),   and\r\n" );
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document.write( "     < 1 + < 2 + < 3 = 180°   (3)     ( as the sum of interior angles of the triangle BPA).\r\n" );
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document.write( "By adding equations (2) and (3), you get  < 1 + < 2 = 90°,  which immediately implies that < 3 = 90°.  \r\n" );
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